What point on the number line is one fourth of the way from the point 0 to the point −3? −0.25 −0.5 −0.75 −1

The shopkeeper would have saved 4756.25 - 4260 = 496.25 shillings.

What is the linear equation?

A linear equation is an algebraic equation of the form y=mx+b. where m is the slope and b is the y-intercept. We have a graph.

Let the cost of one panga be x and one jembe be y.

From the given information, we can form the following equations:

50x + 30y = 4260 --- Equation 1

25x/2 + (30/2 - 5)y = 2970 --- Equation 2

50(1.25)x + 30(0.85)y = total cost from wholesaler B --- Equation 3

Simplifying equation 2:

25x/2 + 10y - 5y = 2970

25x/2 + 5y = 2970

25x + 10y = 5940

Simplifying equation 3:

62.5x + 25.5y = total cost from wholesaler B

To solve for x and y, we can use any method of our choice. For simplicity, we will use elimination:

Multiplying equation 1 by 5:

250x + 150y = 21300 --- Equation 4

Multiplying equation 2 by 2:

25x + 20y = 5940 --- Equation 5

Subtracting equation 5 from equation 4:

225x + 130y = 15360

Substituting the value of y from equation 5:

225x + 130(297 - 2.5x) = 15360

225x + 38610 - 325x = 15360

-100x = -23250

x = 232.5

Substituting the value of x in equation 1:

50(232.5) + 30y = 4260

y = 85

Therefore, the cost of one panga is 232.5 shillings and the cost of one jembe is 85 shillings.

To find out how much the shopkeeper would have saved if he had bought from wholesaler B, we need to calculate the total cost from wholesaler B:

50(1.25)(232.5) + 30(0.85)(85) = 4756.25

The total cost from wholesaler A was 4260 shillings.

Therefore, the shopkeeper would have saved 4756.25 - 4260 = 496.25 shillings.

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Answer:

Step-by-step explanation:

The equation of the line with a slope of 1/6 passes through the point (-6, 5) is y=(1/6)x+6.

What is the equation of a line?

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The equation of line is given by,

y =mx + c

where,

x is the coordinate of the x-axis,

y is the coordinate of the y-axis,

m is the slope of the line, and

c is y-intercept.

Given that a line with a slope of 1/6 passes through the point (-6, 5). Therefore, we can write,

y = mx + c

Substitute the values,

5 = (1/6)(-6) + C

5 = -1 + C

5 + 1 = C

C = 6

Hence, the equation of the line with a slope of 1/6 passes through the point (-6, 5) is y=(1/6)x+6.

Answer:

y = (1/6)x + 7

Step-by-step explanation:

The slope-intercept form of the equation of a line is y = mx + b, where m is the slope and b is the y-intercept.

We know that the line has a slope of 1/6 and passes through the point (-6, 6). To find the y-intercept, we can substitute the values of the point into the equation and solve for b:

y = mx + b

6 = (1/6)(-6) + b

6 = -1 + b

b = 7

Now that we know the slope and y-intercept, we can write the equation of the line in slope-intercept form:

y = (1/6)x + 7

Therefore, the equation of the line in slope-intercept form is y = (1/6)x + 7.

(Please could you kindly mark my answer as brainliest)

The cost of one metre of cotton fabric is £ 7. The cost of one metre polyester fabric is  £ 2.

The ratio of the cost of one metre of polyester fabric to the cost of one metre of cotton fabric is 2: 7.

1 m polyester / 1 m cotton = 2 / 7

1 m cotton = 7/2 × 1 m polyester

for 2 m of polyester fabric, the cost is £ 4

for 1 m of polyester fabric, the cost is £ 4/2 = £ 2

for 1 m of cotton fabric = 7/2 × 1 m polyester fabric = 7/2 × £2 = £ 7

for 1 m of cotton fabric, the cost is £ 7

For 2m, polyester fabric   2 × £2  = £ 4

cotton fabric  2 × £7  = £ 14

For 6m, polyester fabric   6 × £2  = £ 12

cotton fabric  6 × £7  = £ 42

For 8m, polyester fabric   8 × £2  = £ 16

cotton fabric  8 × £7  = £ 56

For 9m, polyester fabric   9 × £2  = £ 18

cotton fabric  9 × £7  = £ 63

The complete table

                           2m        6m      8m      9m

polyester fabric   £ 4       £ 12    £ 16     £ 18

cotton fabric        £ 14     £ 42    £ 56    £ 63

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Answer:

50.2 is the answer as the answer came in point

The possible rational roots of the polynomial are ±1/2, ±1, ±3/2, ±3, ±9/2, ±9 while the actual roots are 1, -3, 3/2

To find the possible rational roots of the polynomial 2x^3 + x^2 - 12x + 9 = 0, we can use the rational root theorem. According to the theorem, if a polynomial with integer coefficients has a rational root p/q (where p and q are integers with no common factors other than 1), then p must be a factor of the constant term (in this case, 9) and q must be a factor of the leading coefficient (in this case, 2).

The factors of 9 are ±1, ±3, and ±9, and the factors of 2 are ±1 and ±2. Therefore, the possible rational roots of the polynomial are:

±1/2, ±1, ±3/2, ±3, ±9/2, ±9

We can now use synthetic division or long division to check which of these possible roots are actual roots of the polynomial. After checking, we find that the real rational root of the polynomial are x = 1, -3, 3/2

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a) Sample Variance = 521.646

b)Two-sided confidence interval for the variance with a confidence level of 0.98 is (5.545, 10029.794).

a) To calculate the sample variance, you will first need to calculate the sample mean. The sample mean is calculated by summing all the observations in the sample and dividing by the number of observations. For this sample, the mean is:

Mean = (32.564 + 7.57 + 21.815 − 13.971 − 15.224) / 5 = 5.168

Next, you will need to calculate the sum of squared deviations from the mean. This is done by subtracting the mean from each observation and squaring the result, and then summing all of the results:

Sum of Squared Deviations = (32.564 - 5.168)^2 + (7.57 - 5.168)^2 + (21.815 - 5.168)^2 + (-13.971 - 5.168)^2 + (-15.224 - 5.168)^2 =
= 1564.939

Finally, you can calculate the sample variance by dividing the sum of squared deviations by the number of observations minus one:

Sample Variance = 1564.939 / (5 - 1) = 521.646

b) To calculate a two-sided confidence interval for the variance with a confidence level of 0.98, you will need to find the critical value from the Chi-squared distribution with a degrees of freedom equal to the number of observations in the sample minus one. For this sample, the degrees of freedom is 4.

The critical value for this degrees of freedom at the given confidence level is 8.37.

The lower bound of the confidence interval is:

Lower bound = (521.646 / 8.37) * (1 - 0.98) = 5.545

The upper bound of the confidence interval is:

Upper bound = (521.646 / 8.37) * (1 + 0.98) = 10029.794

Therefore, the two-sided confidence interval for the variance with a confidence level of 0.98 is (5.545, 10029.794).

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The average speed during a given time interval can be found by calculating the change in distance over the change in time, or (d(b) - d(a))/(b-a). In this case, we can use the given function d(t) = 16t^2 to find the distance at each given time.  It is reasonable to say that this value is the speed of the object at the instant t = 9 because as the time interval approaches zero, the average speed approaches the instantaneous speed at that moment.

For the first time interval, t = a = 9 and t = b = 9.5:
d(a) = 16(9)^2 = 1296
d(b) = 16(9.5)^2 = 1444
Average speed = (1444 - 1296)/(9.5 - 9) = 148/0.5 = 296 ft/s

For the second time interval, t = a = 9 and t = b = 9.1:
d(a) = 16(9)^2 = 1296
d(b) = 16(9.1)^2 = 1324.96
Average speed = (1324.96 - 1296)/(9.1 - 9) = 28.96/0.1 = 289.6 ft/s

For the third time interval, t = a = 9 and t = b = 9.01:
d(a) = 16(9)^2 = 1296
d(b) = 16(9.01)^2 = 1300.9616
Average speed = (1300.9616 - 1296)/(9.01 - 9) = 4.9616/0.01 = 496.16 ft/s

For the fourth time interval, t = a = 9 and t = b = 9.001:
d(a) = 16(9)^2 = 1296
d(b) = 16(9.001)^2 = 1296.288016
Average speed = (1296.288016 - 1296)/(9.001 - 9) = 0.288016/0.001 = 288.016 ft/s

For the fifth time interval, t = a = 9 and t = b = 9.0001:
d(a) = 16(9)^2 = 1296
d(b) = 16(9.0001)^2 = 1296.0288016
Average speed = (1296.0288016 - 1296)/(9.0001 - 9) = 0.0288016/0.0001 = 288.016 ft/s

As the time intervals get smaller and smaller, the average speed approaches 288 ft/s.

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Answer:

6

Step-by-step explanation:

[tex] \frac{2 + 3 + 5 + 7 + 13}{5} = \frac{30}{5} = 6[/tex]

The dairy farm yields 6,158,000 gallons of milk in one year, assuming that each cow produces 2320 gallons per year.

The term "total amount" refers to the complete or full quantity or sum of something. It is the complete amount of something without any deductions or subtractions.

To calculate the total amount of milk the dairy farm yields in one year mathematically, we need to multiply the number of dairy cows by the amount of milk each cow produces in a year.

Let's represent the number of dairy cows as "C" and the amount of milk produced per cow per year as "M". Using this notation, we can write:

Total amount of milk produced in one year = Number of cows x Amount of milk per cow

= C x M

Substituting the given values, we have:

C = 2650 (number of dairy cows)

M = 2320 gallons (amount of milk per cow per year)

Total amount of milk produced in one year = 2650 x 2320

= 6,158,000 gallons

Therefore, the dairy farm yields 6,158,000 gallons of milk in one year, assuming that each cow produces 2320 gallons per year.

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Answer:

6.158 x 10^6

Step-by-step explanation:

Converted answer above to scientific notation.

The given statement "The mode number of TVs per household for both Mrs. Jamison's class and and Mr. Zimmerman's class is 3." is true. The mode of TVs per household is 3 in both Mrs. Jamison's class and and Mr. Zimmerman's class.

Mode is a statistical measure that represents the value that appears most frequently in a dataset. It is one of the measures of central tendency, along with mean and median.

In Mrs. Jamison's class, the mode number of TVs per household is 3 because it appears twice, while all other numbers appear only once. Similarly, in Mr. Zimmerman's class, the mode number of TVs per household is 3 because it appears twice, while all other numbers appear only once.

Therefore, the statement "The mode number of TVs per household for both Mrs. Jamison's class and Mr. Zimmerman's class is 3" is true.

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Answer:

See attached graph for the two functions

y = cos(x)

y = 0.5

Solution set for cos(x) i.e. the values of x for which cos(x) = 0.5 in the interval 0 < x < 2π are
{π/3, 5π/3)

or

{1.05, 5.24}  in decimal

Step-by-step explanation:

I moved the original horizontal up to y = 0.5

The solutions to the two equations are where the two functions intersect

There are two intersection in the interval 0 ≤ x ≤ 2π and are at the points labeled A and B

The two points can be obtained by setting
cos(x) = 0.5 and solving for x

cos(x) = 0.5

=> x = cos⁻¹ (0.5)

= 60° and 300° in the range 0 ≤ x ≤ 2π where 2π = 360°

In terms of π,

Since π radians = 180°, 1° = π/180 radians

60° = π/180 x 60 = π/3 radians

300° =  π/180 x 300 = 5π/3 radians

Therefore the solutions to cos(x) = 0.5 are
x = π/3 and x = 5π/3
The solution set is written as {π/3, 5π/3}

In decimal

π/3 = 1.04719 ≈ 1.05

5π/3 = 5.23598 ≈ 5.24

Solution set in decimal: {1.05, 5.24}

The answer is: Logan's rope is longer than Sam's rope.

A number that represents a part of a whole or a ratio between two quantities, written as a numerator over a denominator. It consists of a numerator (top) and a denominator (bottom) separated by a fraction bar. For example, 1/2 represents one-half of a whole or the ratio of one to two.

Part A:

Brittney's rope is shorter than Sam's rope because it is 4/5 as long as Sam's rope.

Logan's rope is longer than Sam's rope because it is 1 1/4 times as long as Sam's rope.

Holly's rope is equal to Sam's rope because it is 8/8 (which simplifies to 1) as long as Sam's rope.

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The value οf P(2 tails) is 37.5%, the cοrrect οptiοn is B.

Prοbability refers tο a pοssibility that deals with the οccurrence οf randοm events.

The prοbability οf all the events οccurring need tο be 1.

The fοrmula οf prοbability is defined as the ratiο οf a number οf favοurable οutcοmes tο the tοtal number οf οutcοmes.

P(E) = Number οf favοurable οutcοmes / tοtal number οf οutcοmes

We are given that;

The sample space=  {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Nοw,

The sample space fοr tοssing a cοin 3 times is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, where H represents heads and T represents tails.

Tο determine P(2 tails), we need tο cοunt the number οf οutcοmes in which there are 2 tails, and divide that by the tοtal number οf οutcοmes:

Number οf οutcοmes with 2 tails: There are three οutcοmes with 2 tails: TTH, THT, and HTT.

Tοtal number οf οutcοmes: There are eight οutcοmes in tοtal.

P(2 tails) = number οf οutcοmes with 2 tails / tοtal number οf οutcοmes = 3/8 = 0.375, which is equivalent tο 37.5%.

Therefοre, the prοbability the answer will be 37.5%.

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Answer: B. A right circular Cylinder

Step-by-step explanation:

By solving the equation -3x = 348, we find that the value of x is -116.

The property of equality that could be used to solve -3x = 348 is the multiplication property of equality, which states that if we multiply both sides of an equation by the same non-zero number, the equation remains equivalent. In this case, we can divide both sides of the equation by -3 to isolate x and solve for it.

Using the multiplication property of equality, we can multiply both sides by -1/3:

(-1/3) * (-3x) = (-1/3) * 348

Simplifying:

x = -116

Therefore, the solution to the equation -3x = 348 is x = -116.

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1. The finance charge on June 11 is $1.98.

2. The new card balance on June 12th is $341.30

To calculate the finance charge and new card balance, we first need to calculate the average daily balance for the billing cycle.

May 13 Charge toys: $129.79 (balance: $129.79)

May 15 Payment $50 (balance: $79.79)

June 1 Charge clothing $135.95 (balance: $215.74)

June 8 Charge Housewares $37.63 (balance: $253.37)

Billing cycle: May 13 to June 11 (30 days)

Average daily balance:

=  (129.79 x 18) + (79.79 x 16) + (215.74 x 10) + (253.37 x 6) / 30

= $152.49

The Finance charge is computed as:

= Average daily balance * Monthly periodic rate.

= 152.49 x (1.3/100)

= $1.98.

To calculate the new card balance on June 12th, we need to add the finance charge and any new charges to the previous balance and subtract any payments made.

Previous balance (as of June 1st) = $215.74

New charges (since June 1st) = $135.95 + $37.63 = $173.58

Payments made (since May 15th) = $50

The New card balance on June 12th will be:

= $215.74 + $173.58 + $1.98 - $50

= $341.30

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Answer:

y = 4x - 16

Step-by-step explanation:

Parallel lines have the same slope. Find slope using the points (-3, -3) and (-1, 5)

slope = m = (5 - -3) / (-1 - -3) = 8/2 = 4

y = mx + b      find b using the point (4, 0), the x-intercept of the parallel line

0 = 4(4) + b

b = -16

equation of the parallel line:

y = 4x - 16

Tristan has 14 nickels and 7 dimes worth of $1. 40.

This is because 1 nickel is worth 5 cents and 1 dime is worth 10 cents.

Let's use the following variables to represent the number of nickels and dimes Tristan has:

n = number of nickels

d = number of dimes

We know that Tristan has $1.40 worth of nickels and dimes. Each nickel is worth $0.05 and each dime is worth $0.10, so we can write an equation based on their values:

0.05n + 0.1d = 1.4

We also know that Tristan has twice as many nickels as dimes:

n = 2d

We can substitute n = 2d into the first equation and solve for d:

0.05(2d) + 0.1d = 1.4

0.1d + 0.1d = 1.4

0.2d = 1.4

d = 7

So Tristan has 7 dimes. Using n = 2d, we can find the number of nickels:

n = 2(7) = 14

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The Answer:

The answer to the equation that Clare gets is:

X1= -3/2 + 7/2i , X2= -3/2 - 7/2i.

The Explanation:

4x^2+12x+58=0

2x^2+6x+29=0

a=2, b=6, c=29

The line plot shows that the construction company has a number of trucks, each of which is designed to haul different amounts of weight.

If all of these trucks are working at the same time, we can calculate how many tons the trucks can carry in total. In order to do this, we need to look at the range of weight each truck is capable of carrying, and then add all of these numbers together.
In order to determine how many tons the trucks can carry, we need to use the information provided by the line plot. The line plot displays the weight each truck can haul, which is given in pounds. We need to convert the weight in pounds to tons in order to find the total weight that the trucks can carry.

To do this, we can use the following conversion factor:1 ton = 2000 poundsWe can use this conversion factor to convert the weight of each truck from pounds to tons. Once we have done this, we can add up the weights of all the trucks to find the total weight that the trucks can carry. Here are the steps:

Step 1: Convert the weight of each truck from pounds to tons Truck 1: 6,000 pounds ÷ 2,000 pounds/ton = 3 tons Truck 2: 9,000 pounds ÷ 2,000 pounds/ton = 4.5 tons Truck 3: 8,000 pounds ÷ 2,000 pounds/ton = 4 tons Truck 4: 10,000 pounds ÷ 2,000 pounds/ton = 5 tons Truck 5: 11,000 pounds ÷ 2,000 pounds/ton = 5.5 tons Truck 6: 9,500 pounds ÷ 2,000 pounds/ton = 4.75 tons

Step 2: Add up the weights of all the trucks3 + 4.5 + 4 + 5 + 5.5 + 4.75 = 26.75 tons. The trucks can carry 26.75 tons in total.

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Answer: Domain: All x-values

Range: All y-values

Step-by-step explanation:

Answer:-14a-4

Step-by-step explanation:

Answer: Expanding the expression, we get:

-7(2a - 1) - 11 = -14a + 7 - 11

Combining like terms, we get:

-14a - 4

Therefore, the simplified expression is -14a - 4.

Enjoy!

hope it helps you

option d

Answer:

|a × [ob + (1 - o)a]| = √(7 - 8o(a · o) - 8(a · o)^2)

where k = 7 - 8o(a · o) - 8(a · o)^2.

Step-by-step explanation:

Given the position vectors of two points A and B as a and b respectively, where b is a unit vector, and the magnitude of a is twice that of b, we are asked to show that:

|a × [ob + (1-o)a]| = √k,

where k is a constant to be determined.

We can begin by expanding the vector inside the cross product:

ob + (1 - o)a = ob + a - oa

Since b is a unit vector, we can write:

ob = b - o

Substituting this into the previous equation, we get:

ob + (1 - o)a = b - o + a - oa = b + (1 - o)a - oa

Next, we can use the vector cross product formula:

|a × b| = |a||b|sinθ

where θ is the angle between a and b.

We are given that the angle between a and b is 60°, so we can substitute this value into the formula:

|a × b| = |a||b|sin60° = (2|b|)(1)(√3/2) = √3

Now we can calculate the cross product of a and the vector we just derived:

a × [ob + (1 - o)a] = a × (b + (1 - o)a - oa)

= a × (b + a - oa)

= a × b + a × a - a × oa

Since b is a unit vector, we know that a × b is a vector perpendicular to both a and b, and therefore perpendicular to the plane containing a and b. The vector a × a is 0 since the cross product of a vector with itself is 0. Finally, we can use the vector triple product to simplify a × oa:

a × oa = (a · a)o - (a · o)a = |a|^2 o - (a · o)a

Since |a| is twice |b|, we have:

|a|^2 = 4|b|^2 = 4

Substituting this back in, we get:

a × oa = 4o - (a · o)a

Putting it all together, we have:

a × [ob + (1 - o)a] = a × b + 4o - (a · o)a

Now we can take the magnitude squared of both sides:

|a × [ob + (1 - o)a]|^2 = (a × b + 4o - (a · o)a) · (a × b + 4o - (a · o)a)

Expanding the dot product, we get:

|a × [ob + (1 - o)a]|^2 = |a × b|^2 + 16o^2 + |a|^2(o · o) - 8o(a · o)b + 8(a · o)(a × b) - 2(a · o)^2|a|^2

Substituting the values we derived earlier, we get:

|a × [ob + (1 - o)a]|^2 = 3 + 16o^2 + 4(o · o) - 8o(a · o) + 0 - 2(a · o)^2(4)

= 7 - 8o(a · o) - 8(a · o)^2

Now we need to find the value of k such that the left-hand side equals k:

|a × [ob + (1 - o)a]|^2 = k

Using the vector triple product again, we can simplify the left-hand side as:

|a × [ob + (1 - o)a]|^2 = |a|^2|ob + (1 - o)a|^2 - ((a · [ob + (1 - o)a])^2)

Since we know that the magnitude of a is twice that of b, we have:

|a|^2 = 4|b|^2 = 4

Substituting this back in, we get:

|a × [ob + (1 - o)a]|^2 = 4|ob + (1 - o)a|^2 - ((a · [ob + (1 - o)a])^2)

Now we can substitute the expanded expression for ob + (1 - o)a:

|a × [ob + (1 - o)a]|^2 = 4|b + (1 - o)a|^2 - ((a · [b + (1 - o)a - oa])^2)

= 4|b|^2 + 8|b|(1 - o)(a · b) + 4(1 - o)^2|a|^2 - ((a · b + (1 - o)(a · b) - (a · o)(a · b))^2)

= 4 + 8(1 - o)(a · b) + 4(1 - o)^2(4) - ((a · b + (1 - o)(a · b) - (a · o)(a · b))^2)

= 28 - 8o(a · b) - 8(a · o)^2

Substituting this back into the previous equation, we get:

28 - 8o(a · b) - 8(a · o)^2 = k

Therefore, we have:

|a × [ob + (1 - o)a]| = √(28 - 8o(a · b) - 8(a · o)^2)  and

k = 28 - 8o(a · b) - 8(a · o)^2

Hope this helps! Sorry if it's wrong! If you need more help, ask me! :]

Answer:

Miguel's account: M(x) = 320 + 85x

Kala's account: K(x) = 820 + 35x

b) To find when the two accounts would have the same amount of money, we need to set M(x) equal to K(x) and solve for x:

320 + 85x = 820 + 35x

Simplifying the equation, we get:

50x = 500

Dividing both sides by 50, we get:

x = 10

Therefore, the two accounts would have the same amount of money after 10 months.

The given quadrilateral is a parallelοgram and a kite.

Quadrilaterals are pοlygοns that have fοur sides, fοur vertices, and fοur angles. They are twο-dimensiοnal shapes that can be classified based οn their prοperties, such as the lengths οf their sides, the measures οf their angles, and the presence οf parallel sides οr right angles. Sοme cοmmοn types οf quadrilaterals include:

Nοw,

Tο identify the type οf quadrilateral fοrmed by the vertices T(-3,-3), U(4, 4), V(0, 6), and W(-5, 1), we need tο first find the lengths οf the sides and the slοpes οf the segments cοnnecting the vertices.

Using the distance fοrmula, we get:

[tex]TU = \sqrt{[(4 - (-3))^2 + (4 - (-3))^2]} = \sqrt {[7^2 + 7^2]} = \sqrt{(98)[/tex]

[tex]UV = \sqrt{[(0 - 4)^2+ (6 - 4)^2]} = \sqrt{[(-4)^2 + 2^2]} = \sqrt{(20)[/tex]

[tex]VW = \sqrt{[(-5 - 0)^2 + (1 - 6)^2]}= \sqrt{[(-5)^2+ (-5)^2]} = \sqrt{(50)[/tex]

[tex]WT = \sqrt{[(-5 - (-3))^2+ (1 - (-3))^2]} = \sqrt{[(-2)^2 + 4^2]} = \sqrt{(20)[/tex]

Next, we can find the slοpes οf the segments:

TU: m = (4 - (-3))/(4 - (-3)) = 1

UV: m = (6 - 4)/(0 - 4) = -1/2

VW: m = (1 - 6)/(-5 - 0) = 1

WT: m = (1 - (-3))/(-5 - (-3)) = -1/2

Nοw we can use these measurements tο identify the type οf quadrilateral:

Oppοsite sides are parallel: VW and TU have slοpes οf 1 and slοpes οf -1 respectively. Therefοre, the quadrilateral is a parallelοgram.

Twο adjacent sides are cοngruent: TU and UV have lengths οf sqrt(98) and sqrt(20) respectively. Therefοre, the quadrilateral is nοt a rhοmbus.

Diagοnals bisect each οther: The diagοnals TV and UW intersect at (2, 1.5), which is the midpοint οf bοth diagοnals. Therefοre, the quadrilateral is a parallelοgram.

One pair οf οppοsite sides are perpendicular: The slοpes οf UV and WT are -1/2, and the prοduct οf their slοpes is -1. Therefοre, the quadrilateral is a kite.

All sides are cοngruent: The lengths οf the sides are nοt all equal. Therefοre, the quadrilateral is nοt a square.

Thus, the quadrilateral fοrmed by the given vertices is a parallelοgram and a kite.

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Answer:

Hope this helps :)

Step-by-step explanation:

Because x is an absolute value, the value of y is always greater than or equal to zero. I attached the graph below. As you'll see, when x is a negative number, it was the same value as the positive of that value. When x = 1 or x = -1, y = 5.

Answer:

y = -9

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

since 2y is adding to positive 6 we take 6 to the right side where there its like term 12. we subtract 6 from 12 which is giving us 6. we are remaining with 2y=6 we divide both sides by 2 giving us y=3

By deducting the value of the shortest distance from the value of the longest distance on the line plot, it is possible to determine the difference between the longest and shortest distances a student has walked to get to school.

We must look at the provided line plot to ascertain the difference between the longest and shortest distances a student walks to get to school. Ten pupils were tracked across various distances using a line plot. The location of each student is indicated by a "X" on the map.

Just looking for the X with the highest and lowest frequency will yield the longest and shortest lengths. According to the line plot, the distances at which Xs occur most frequently are 2 miles away and 0.5 miles away, respectively. As a result, there is a 1.5 mile discrepancy between the student's maximum walking distance (2 miles) and their shortest walking distance (0.5 miles).

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1. Quadratic monomials.

2.  Biquadratic five-term polynomials.

3. Quadratic trinomials.

4. x³ + 3x² - 5x - 4

5. -[tex]x^{5}[/tex] + 4[tex]x^{4}[/tex] +2x³ + 2x - 7

6. - x² + 5x + 9

7. y² - 3y - 9

8. 5(x³ + x)

9. 2x² + 2x -5

Algebraic expressions called polynomials only have non-negative integer powers for their variables. A polynomial is, for instance, 5x² - x + 1. The polynomial 3x³ + 4x + 5/x + 6[tex]x^{3/2}[/tex] is not a polynomial since one of the powers of "x" is a fraction and the other is negative.

Expressions with one or more terms that have a non-zero coefficient are called polynomials. Variables, exponents, and constants make up polynomial terms. The "leading term" refers to the first term of the polynomial in standard form.

Here in the given question,

We can see the highest degree of the variable and we can determine the name of each polynomial.

Likewise, we can just arrange the expressions as per the highest value of the power of the variable.

And simplify the expression by adding or subtracting the like terms.

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