Answer:
x = - [tex]\frac{5}{2}[/tex] , x = - [tex]\frac{4}{3}[/tex]
Step-by-step explanation:
6x² + 23x + 20 = 0 ( factorise the left side )
consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 6 × 20 = 120 and sum = 23
the factors are + 8 and + 15
use these factors to split the x- term
6x² + 8x + 15x + 20 = 0 ( factor the first/second and third/fourth terms )
2x(3x + 4) + 5(3x + 4) = 0 ← factor out (3x + 4) from each term
(3x + 4)(2x + 5) = 0
equate each factor to zero and solve for x
2x + 5 = 0 ⇒ 2x = - 5 ⇒ x = - [tex]\frac{5}{2}[/tex]
3x + 4 = 0 ⇒ 3x = - 4 ⇒ x = - [tex]\frac{4}{3}[/tex]
Finding supplementary and complementary angles (a) An angle measures 50°. What is the measure of its complement? (b) An angle measures 135°. What is the measure of its supplement? measure of the complement: measure of the supplement: 0 0 O X ?
SOLUTION
(a) Complementary angles are angles that add up to 90 degrees. So the angle that will complement 50 degrees will add to it to get 90. Let the angle be x, we have
[tex]\begin{gathered} 50\degree+x\degree=90\degree \\ 50+x=90 \\ x=90-50 \\ x=40\degree \end{gathered}[/tex]Hence the measure of the compelement is 40 degrees
(b) Supplementary angles are angles that add up to 180 degrees. So the angle that will supplement 135 degree will add to it to make it 180 degrees. Let this angle be y, so we have
[tex]\begin{gathered} 135\degree+y\degree=180\degree \\ y=180-135 \\ y=45\degree \end{gathered}[/tex]Hence measure of the supplement is 45 degrees
Identify the values of a, b, and c for the quadratic equation given:y=-x2 +9a =b =C=
Question:
Solution:
A quadratic Equation in Standard Form is given by the following formula:
[tex]ax^2+bx\text{ + c = 0}[/tex]now, the given equation is
[tex]y=-x^2+9[/tex]this is equivalent to:
[tex]f(x)=-x^2+9[/tex]According to the Quadratic Equation in Standard Form, we can conclude that
[tex]a\text{ = -1}[/tex][tex]b\text{ = 0}[/tex]
and
[tex]c\text{ = 9}[/tex]Hey everybody! Can somebody help me solve this problem? I don't need a big explanation just the answer and a brief explanation on how you get it! Look at photo for problem.
Given the ordered pairs:
(-12, -16), (-3, -4), (0, 0), (9, 12)
Let's say that the first coordinate corresponds to x, and the second one corresponds to y. Then, the constant of variation k relates x and y as:
[tex]y=k\cdot x[/tex]Using the ordered pairs:
[tex]\begin{gathered} -16=-12k\Rightarrow k=\frac{4}{3} \\ -4=-3k\Rightarrow k=\frac{4}{3} \\ 0=0\cdot k\text{ (this means that it is correct)} \\ 12=9k\Rightarrow k=\frac{4}{3} \end{gathered}[/tex]We conclude that the constant of variation is:
[tex]k=\frac{4}{3}[/tex]what is the area if one of the triangular side of the figure?
Compound Shape
The shape of the figure attached consists on four triangles and one square.
The base of each triangle is B=12 cm and the height is H=10 cm, thus the area is:
[tex]A_t=\frac{BH}{2}[/tex]Calculating:
[tex]A_t=\frac{12\cdot10}{2}=60[/tex]The area of each triangle is 60 square cm.
Now for the square of a side length of L=12.
The area of a square of side length a, is:
[tex]A_s=a^2[/tex]Calculate the area of the square:
[tex]A_s=12^2=144[/tex]The total surface area is:
A = 60*4 + 144
A= 240 + 144
A = 384 square cm
You roll a die. What is the probability that you’ll get a number less than 3?0.3330.50.6670.75
Recall that the numbers in a die are 1,2,3,4,5,6.
[tex]S=\mleft\lbrace1,2,3,4,5,6\mright\rbrace[/tex]Hence the number of possible outcomes is 6.
[tex]n(S)=6[/tex]We need a number less than 3. Let A be this event.
[tex]A=\mleft\lbrace1,2\mright\rbrace[/tex]The favorable outcome is 2.
[tex]n(A)=\mleft\lbrace1,2\mright\rbrace[/tex]Since there are 1,2 less than 3 in a die.
[tex]P(A)=\frac{Favourable\text{ outcomes}}{\text{Total outcomes}}=\frac{n(A)}{n(S)}[/tex]Substitute n(A)=2 and n(S)=6, we get
[tex]P(A)=\frac{2}{6}=\frac{1}{3}=0.333[/tex]Hence the required probability is 0.333.
A website recorded the number y of referrals it received from social media websites over a 10-year period. The results can be modeled by y = 2500(1.50), where t is the year and 0 ≤ t ≤9.Interpret the values of a and b in this situation.O a represents the number of referrals after 9 years; b represents the growth factor of the number of referrals each year.a represents the number of referrals it received at the start of the model; b represents the decay factor of the number of referralseach year.O a represents the number of referrals after 9 years; b represents the decay factor of the number of referrals each year.a represents the number of referrals it received at the start of the model; b represents the growth factor of the number ofreferrals each year.What is the annual percent increase?The annual percent increase is%.
In this problem
we have an exponential growth function
[tex]y=2500(1.50)^t[/tex]where
a=2500 ----> initial value of the number of referrals at the start of the model
b=1.50 ---> the base of the exponential function (growth factor of the number of referrals each year
therefore
The answer Part 1 is the last option
Part 2
Find out the annual percent increase
we know that
b=1+r
b=1.50
1.50=1+r
r=1.50-1
r=0.50
therefore
r=50%
The annual percent increase is 50%3. An equation that crosses the y-axis at -5 and crosses the x-axis at 24. An equation that crosses the y-axis at -5 and crosses the x-axis at -65. An equation that crosses the y-axis at -5 and crosses the point (2,3)
We need to find the equation of the line which:
• crosses the y-axis at -5
,• crosses the x-axis at 2
The y-axis cutting point is (0,-5)
The x-axis cutting point is (2,0)
The equation of line is:
[tex]y=mx+b[/tex]Where m is the slope and b is the y-axis cutting point
m is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Where
y_2 = 0
y_1 = -5
x_2 = 2
x_1 = 0
So, slope is:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{0--5}{2-0}=\frac{0+5}{2}=\frac{5}{2}[/tex]We got m, we also know b.
The y cutting point is -5, so b = -5
The equation is:
[tex]y=\frac{5}{2}x-5[/tex]The graph would look like:
More clear version:
Convert the function p(x) = 2(x – 4)(x + 3)
Expanding the expression,
[tex]\begin{gathered} p(x)=2(x-4)(x+3) \\ \rightarrow p(x)=2(x^2+3x-4x-12) \\ \rightarrow p(x)=2(x^2-x-12) \\ \rightarrow p(x)=2x^2-2x-24 \end{gathered}[/tex]We get that:
[tex]p(x)=2x^2-2x-24[/tex]The displacement (in meters) of a particle moving in a straight line is given by s = t^2 - 9t + 15,where t is measured in seconds.(A)(i) Find the average velocity over the time interval [3,4].Average Velocity = ___ meters per second(ii) Find the average velocity over the time interval [3.5,4].Average Velocity=____meters per second(iii) Find the average velocity over the time interval [4,5].Average Velocity= ____meters per second(iv) Find the average velocity over the time interval (4,4.5] Average Velocity = ____meters per.(B) Find the instantaneous velocity when t=4.Instantaneous velocity= ____ meters per second.
Given
The displacement (in meters) of a particle moving in a straight line is given by s = t^2 - 9t + 15,
The graph models the heights, in feet, of two objectsdropped from different heights after x seconds.Which equation represents g(x) as a transformation off(x)?y45+O g(x) = f(x) -5O g(x) = f(x-5)O g(x) = f(x) + 5O g(x) = f(x + 5)40+35y = f(x)30-25+20-15+10+5+ly = g(x)0.5 1.0 1.5 20
For this problem we know that y=f(x) at x=0 is 5 units above y=g(x). So then the best solution for this case it seems to be:
[tex]f(x)=g(x)+5[/tex]And solving for g(x) we got:
[tex]g(x)=f(x)-5[/tex]One angle measures 140°, and another angle measures (5k + 85)°. If the angles are vertical angles, determine the value of k.
The value of k when one angle measures 140°, and another angle measures (5k + 85)° and if the angles are vertical angles is 11.
What is vertical angles?
Vertical angles are angles opposite each other where two lines cross.
Note: Vertical angles are equal.
To calculate the value of k, we use the principle of vertical angle
From the question,
140 = (5k+85)°Solve for k
5k = (140-85)5k = 55Divide both side by the coefficient of k (5)
5k/5 = 55/5k = 11Hence, the value of k is 11.
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Graph the line with the given slope m and y-intercept b.
m = 4, b = -5
Answer:
Step-by-step explanation:
What we know:
m = 4, b = -5
y = mx + b where m is the gradient/slope and b is the y-intercept
Substitute m and b values:
y = 4x + -5 which is the same as y = 4x - 5
Substitute all x values to find y coordinate:
When x = -7, y = (4 x -7) - 5 = -33
When x = -6, y = (4 x -6) - 5 = -29
When x = -5, y = …
Continue for all x values
!!PLEASE HELP IMMEDIATELY!!
Solve the inequality
-1/3x - 12 > 21 or -6x + 10 < -2
x < ? or x > ?
solve for both
Answer:
x < 2 or x > -11
Step-by-step explanation:
b. 1. add 12 to both sides to get -1/3x > 33
2. multiply by -3/1 to both sides to get x > -11
a. 1) subtract 10 to both sides
2) divide by -6 to both sides
5 Which equations have the same value of x as 6 2 3 -9? Select three options. -9(6) 5x+4=-54 5x+4=-9 5x=-13 5X=-58
The given equation is-
[tex]\frac{5}{6}x+\frac{2}{3}=-9[/tex]If we multiply the equation by 6, we would have the same value for the variable x since we are multiplying the same number on each side. So, the second choice is an equivalent equation to the given one.
Let's multiply by 6.
[tex]\begin{gathered} 6\cdot\frac{5}{6}x+6\cdot\frac{2}{3}=-9\cdot6 \\ 5x+4=-54 \end{gathered}[/tex]So, the third expression is also an equivalent expression.
Then, let's subtract 4 on each side.
[tex]\begin{gathered} 5x+4-4=-54-4 \\ 5x=-58 \end{gathered}[/tex]The last choice is also an equivalent expression.
Therefore, the right choices are 2, 3, and 6.What is the slope of the line that passes through the points (6,-10) and (3,-13)? Write in simplist form
Use the slope formula to find the slope of a line that goes through two points:
[tex]\begin{gathered} \text{Coordinates of two points}\rightarrow\text{ }(x_1,y_1),(x_2,y_2) \\ \text{Slope of a line through those points}\rightarrow m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]Substitute the coordinates (6,-10) and (3,-13) into the slope formula:
[tex]\begin{gathered} m=\frac{(-13)-(-10)}{(3)-(6)} \\ =\frac{-13+10}{3-6} \\ =\frac{-3}{-3} \\ =1 \end{gathered}[/tex]Therefore, the slope of a line that passes through those points, is 1.
Maria is at the top of a cliff and sees a seal in the water. If the cliff is 40 feet above the water, Marla's eye-level is 5.5 feet, and the angle of depression is 52°, what is the horizontal distance from the seal to the cliff, tothe nearest foot?
SOLUTION
Let us make a diagram to interpret the question
from the diagram above, we can make the right-angle triangle as follows
So we can use SOHCAHTOA to solve this. The opposite side to the angle 52 degrees is 45.5 ft, this is gotten by adding the height of the cliff to Maria's height from her feet to her eyes.
The adjacent side is d, that is the distance from the seal to the cliff, so we have
[tex]\begin{gathered} TOA\text{ tan}\theta\text{ = }\frac{opposite}{adjacent} \\ tan52\degree=\frac{45.5}{d} \\ cross\text{ multiply, we have } \\ tan52\degree d=45.5 \\ d=\frac{45.5}{tan52} \\ d=35.54849 \end{gathered}[/tex]Hence the answer is 36 foot to the nearest foot
Find the interval in which the following quadratic is decreasing.
The quadratic is decreasing in the interval in which the y values decrease with the increase in x values.
In the interval, (-∞, 0), the y values decrease with increase in x values.
Hence, the quadratic is decreasing in the interval (-∞, 0),
The area in square millimeters of a wound has decreased by the same percentage every day since it began to heal. The table shows the wound's area at the end of each day.
Given the table showing the number of days since wound began to heal and area of wound in square millimeters
To determine the statement that are correct from the option provided
From the table shown it can be seen that as the day increases by 1, the area of wound in square millimeters decreases by a common ratio of
[tex]\frac{20}{25}=\frac{16}{20}=\frac{12.8}{16}=\frac{10.24}{12.8}=0.8[/tex]Suppose that an expression to represent the area of wound is
[tex]ab^c[/tex]The modelled expression from the table is
[tex]\begin{gathered} a=25 \\ b=0.8 \\ c=n-1 \\ \text{Therefore, we have} \\ 25(0.8^{n-1}) \end{gathered}[/tex]Let us use the modelled expression to verify each of the given conditions
The modelled expression can be simplified as shown below:
[tex]\begin{gathered} 25(0.8^{n-1}) \\ \text{Note},\text{ using indices rule} \\ \frac{a^n}{a}=a^{n-1} \\ \text{Therefore:} \\ 0.8^{n-1}=\frac{0.8^n}{0.8} \end{gathered}[/tex]Then, we have the modelled expression becomes
[tex]25(0.8^{n-1})=25\times\frac{0.8^n}{0.8}=\frac{25}{0.8}\times0.8^n=31.25(0.8^n)[/tex]From the two modelled expression we can see that
[tex]\begin{gathered} \text{when:} \\ c=n-1,a=25,b=0.8 \\ c=n,a=31.25,b=0.8 \end{gathered}[/tex]Then we can conclude that the two conditions that are true from the options are
If the value of c = n, the value of a is 31.25, and
If the value of c = n, the value of b is 0.8
solve 2x^2+5x-3>0 quadratic inequalities
The solution set of the inequality 2 · x² + 5 · x - 3 > 0 is (- ∞, - 3) ∪ (1 / 2, + ∞).
How to solve a quadratic inequality
Herein we find a quadratic inequality, whose solution set can be found by factoring the expression and determine the interval where the expression is greater than zero. Initially, we use the quadratic formula to determine the roots of the quadratic function:
2 · x² + 5 · x - 3 = 0
x₁₂ = [- 5 ± √[5² - 4 · 2 · (- 3)]] / (2 · 2)
x₁₂ = (- 5 ± 7) / 4
x₁ = 1 / 2, x₂ = - 3
Then, the factored form of the inequality is:
(x - 1 / 2) · (x + 3) > 0
In accordance with the law of signs, we must look for that intervals such that: (i) (x - 1 / 2) > 0, (ii) (x + 3) > 0, (ii) (x - 1 / 2) < 0, (x + 3) < 0. Then, the solution set of the quadratic inequality is:
Inequality form - x > 1 / 2 ∨ x < - 3
Interval form - (- ∞, - 3) ∪ (1 / 2, + ∞)
The solution set of the inequality is (- ∞, - 3) ∪ (1 / 2, + ∞).
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Two functions, function A and function B, are shown below:Function Axy714816918Which statement best compares the rate of change of the two functions?The rate of change of both functions is 2.The rate of change of both functions is 3.The rate of change of function A is greater than the rate of change of function B.The rate of change of function B is greater than the rate of change of function A.
Answer
The rate of change of both functions is 2.
Explanation
To know the statement that best compares the rate of change of the two functions, we need to first calculate the rate of change for each function.
Rate of change of function A
Using x₁ = 7, y₁ = 14, x₂ = 8 and y₂ = 16
Rate of change = Δy/Δx
Δy = (y₂ - y₁) = 16 - 14 = 2
Δx = (x₂ - x₁) = 8 - 7 = 1
⇒ Rate of change = 2/1 = 2
Rate of change of function B
From the graph
Using coordinate x₁ = 2, y₁ = 4, x₂ = 3 and y₂ = 6
Rate of change = Δy/Δx
Δy = (y₂ - y₁) = 6 - 4 = 2
Δx = (x₂ - x₁) = 3 - 2 = 1
⇒ Rate of change = 2/1 = 2
Since the rate of both functions are the same (2), then the statement that best compares the rate of change of the two functions in the options given is "The rate of change of both functions is 2"
Find the sum of the arithmetic series 31+37 +43 +49 +... where n=8,OA. 416B. 1668OC. 832D. 834Reset Selection
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given details
[tex]\begin{gathered} a_1=31 \\ n=8 \\ d=37-31=6 \end{gathered}[/tex]STEP 2: Write the formula for finding sum of arithmetic series
STEP 3: Find the sum of the series
By substitution,
[tex]\begin{gathered} S_8=\frac{8}{2}[2(31)+(8-1)6] \\ S_8=4(62+42) \\ S_8=4(104)=416 \end{gathered}[/tex]Hence, the sum is 416
Exercise 2 Find a formula for Y in terms of X
Given:
y is inversely proportional to square of x.
The equation is written as,
[tex]\begin{gathered} y\propto\frac{1}{x^2} \\ y=\frac{c}{x^2}\ldots\ldots\ldots c\text{ is constant} \end{gathered}[/tex]Also y = 0.25 when x = 5.
[tex]\begin{gathered} y=\frac{c}{x^2} \\ 0.25=\frac{c}{5^2} \\ 25\times0.25=c \\ c=\frac{25}{4} \end{gathered}[/tex]So, the equation of y interms of x is,
[tex]y=\frac{25}{4x^2}[/tex]When x increases,
[tex]\begin{gathered} \lim _{x\to\infty}y=\lim _{x\to\infty}(\frac{25}{4x^2}) \\ =\frac{25}{4}\lim _{x\to\infty}(\frac{1}{x^2}) \\ =0 \end{gathered}[/tex]Hence, the value of x increases then y decreases.
Suppose you are completely locked out outside of your house. You remember that you left your bedroom window, which is 12 feet above the ground, unlocked from the inside (meaning you can climb up the window if you have a ladder, which you do!). You go to your garage, grab the 20 ft ladder and place it so that it reaches exactly to your bedroom window. What is the angle of elevation needed to reach your window? How far away will the bottom of the ladder be from your house?
A diagram of the situation is shown below:
In order to determine the angle of elevation x, use the sine function, as follow:
sin x = opposite side/hypotenuse
the opposite side is the distance from the ground to the wi
2. The product of two consecutive odd numbers is 143. Find the numbers. (Hint: If the first odd number is x, what is the next odd number?)
Step-by-step explanation:
we have the 2 numbers x and (x+2).
x × (x + 2) = 143
x² + 2x = 143
x² + 2x - 143 = 0
the general solution to such a quadratic equation
ax² + bx + c = 0
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case this is
x = (-2 ± sqrt(2² - 4×1×-143))/(2×1) =
= (-2 ± sqrt(4 + 572))/2 = (-2 ± sqrt(576))/2 =
= (-2 ± 24)/2 = (-1 ± 12)
x1 = -1 + 12 = 11
x2 = -1 - 12 = -13
so, we have 2 solutions : 11 and 13, -13 and -11
11× 13 = 143
-11×-13 = 143
A rectangular garden plot measure 3.1 meters by 5.6 meters as shown Find the area of the garden in square meters
Given:
Length(l) of the garden is 3.1 meters
Width(w) of the rectangular garden is 5.6 meters
[tex]\begin{gathered} \text{Area of the garden=}l\times w \\ =3.1\times5.6 \\ =17.36 \end{gathered}[/tex]Area of the garden is 17.36 square meters.
Hello I I am confused because their are two different letters.
Let's begin by listing out the information given to us:
Line AB is parallel to Line CD; this implies that the angle formed by the two lines are right angles (90 degrees)
E is the intersecting point of both lines AB & CD (figure attached)
Let us put this into its mathematical form:
[tex]\begin{gathered} m\angle AED=(6x-24)=90^{\circ} \\ 6x-24=90\Rightarrow6x=90+24 \\ 6x=114\Rightarrow x=19 \\ x=19 \\ m\angle CEB=(4y+32)=90^{\circ} \\ 4y+32=90\Rightarrow4y=90-32 \\ 4y=58\Rightarrow y=17 \\ y=17 \end{gathered}[/tex]Find the minimum weight resistance possible for A 230 pound man
Hello there. To find this minimum weight resistance, we need to convert the percentage value to decimals and multiply it by the weight of the person.
8% converted to decimals is equal to 0.08.
Now, multiply it by the weight of the 230 pound man
0.08 * 230 = 18.4 pounds
This is the minimum weight resistance this U gym offers to the customers.
15 points?Solve for A 5/A = P A = ???? That’s all it saysPlease state what A is.
You are selling drinks at the carnival to raise money for your club. You sell lemonadefor $6 for 2 cups and orange drinks for $9 for 3 cups. Your sales totaled $240. Let xbe the number of cups of lemonade and y be the number of orange drinks. Write anyequation in standard form for the relationship above.
Let x be the number of cups of lemonade sold, and y the number of cups of orange drinks sold, then we can set the following equation:
[tex]6(\frac{x}{2})+9(\frac{y}{3})=240.[/tex]Now, recall that the standard form of a linear equation is:
[tex]Ax+By=C,[/tex]Where, A≥0, B and C are integers.
Simplifying the first equation, we get:
[tex]3x+3y=240.[/tex]Answer:
[tex]3x+3y=240.[/tex]An 80% confidence interval for a proportion is found to be (0.27, 0.33). Whatis the sample proportion?
Step 1
Given;
Step 2
When repeated random samples of a certain size n are taken from a population of values for a categorical variable, the mean of all sample proportions equals the population percentage (p).
[tex]\begin{gathered} Sample\text{ proportion=}\hat{p} \\ \hat{p}\pm margin\text{ error=cofidence interval} \end{gathered}[/tex]Thus;
[tex]\begin{gathered} Let\text{ }\hat{p}=x \\ Margin\text{ of error=y} \\ x-y=0.27 \\ x+y=0.33 \end{gathered}[/tex]checking properly, the sample proportion =0.30, because
[tex]\begin{gathered} 0.30-0.03=0.27 \\ 0.30+0.03=0.33 \end{gathered}[/tex]Answer; Option D
[tex]0.30[/tex]