L is the midpoint of JM. K is the midpoint of JL. JL = 15. What is thelength of KM?

Answers

Answer 1

We can draw the situation as:

So, if L is the midpoint of JM, JL is equal to LM and JM is equal to 2 times JL

JL = LM

JL + LM = JM

JL + JL = JM

2JL = JM

We can calculate JM using JL as:

2JL = JM

2*15 = JM

30 = JM

Then, if K is the midpoint of JL, JK is equal to KL and JL is 2 times JK

JL = 2JK

Replacing JL by 15, we get:

15 = 2JK

15/2 = JK

7.5 = JK

Finally, KM can be calculated as:

KM = JM - JK

KM = 30 - 7.5

KM = 22.5

Answer: 22.5

L Is The Midpoint Of JM. K Is The Midpoint Of JL. JL = 15. What Is Thelength Of KM?

Related Questions

Rewrite the fraction with a rational denominator:
[tex]\frac{1}{\sqrt{5} +\sqrt{3} -1}[/tex]
Give me a clear and concise explanation (Step by step)
I will report you if you don't explain

Answers

The expression with rational denominator is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]

How to rewrite the fraction?

From the question, the fraction is given as

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1}[/tex]

To rewrite the fraction with a rational denominator, we simply rationalize the fraction

When the fraction is rationalized, we have the following equation

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{1}{\sqrt 5 + \sqrt{3} - 1} \times \frac{\sqrt 5 - \sqrt{3} + 1}{\sqrt 5 - \sqrt{3} + 1}[/tex]

Evaluate the products in the above equation

So, we have

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

This gives

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{5 - 10 - 2\sqrt 3}[/tex]

So, we have

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3}[/tex]

Rationalize again

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3} \times \frac{- 5+2\sqrt 3}{- 5 +2\sqrt 3}[/tex]

This gives

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

So, we have

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{25 -12}[/tex]

Evaluate

[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]

Hence, the expression is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]

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Surface are of the wood cube precision =0.00The weight of the woo cube precision =0.00 The volume was 42.87 in3

Answers

Given:

The volume of the cube is 42.87 cubic inches.

The volume of a cube is given as,

[tex]\begin{gathered} V=s^3 \\ 42.87=s^3 \\ \Rightarrow s=3.5 \end{gathered}[/tex]

The surface area of a cube is,

[tex]\begin{gathered} SA=6s^2 \\ SA=6\cdot(3.5)^2 \\ SA=73.5 \end{gathered}[/tex]

Answer: the surface area is 73.5 square inches ( approximately)

what is an identityA) an identity is a false equation relating to a mathematical expression to a real numberB) an identity is a true equation relating to a mathematical expression to a real numberC) an identity is a true equation relating one mathematical expression to another expressionD) an identity is a false equation relating to one mathematical expression to another expression

Answers

The right answer is C

5-3/2x>1/3what is x?

Answers

Coco, this is the solution to the inequality:

5 - 3x/2 ≥ 1/3

Subtracting 5 at both sides:

5 - 3x/2 - 5 ≥ 1/3 - 5

-3x/2 ≥ 1/3 - 15/3

-3x/2 ≥ -14/3

LCD (Least Common Denominator) between 2 and 3 : 6

-9x/6 ≥ -28/6

Dividing by -9/6 at both sides:

-9x/6 / -9/6 ≥ -28/6 / -9/6

x ≥ 28/9

In consequence, the correct answer is C. x ≥ 28/9

Gloria's teacher asks her to draw a triangle with a 90° angle and a 42° angle.How many unique triangles can Gloria draw that meet her teacher's requirements?AOne unique triangle can be drawn because the third angle must measure 48º.BNo unique triangle can be drawn because the teacher only gave the measures of two angles.СInfinitely many unique triangles can be drawn because the side lengths of the triangles can be different sizes.DThere is not enough information to determine how many unique triangles can be drawn.

Answers

SOLUTION

Sum of angles in a triangle must be equal to 180°

So, since one of the angle measures 90° and the other is 42°, then

90 + 42 + y = 180°, where y is the third angle

So, 132 + y = 180

y = 180 - 132 = 48°.

Therefore, one unique triangle can be drawn because the third angle must measure 48º.

Option A is the correct answer.

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Z2
Find the midpoint m of z₁ = (9+7i) and Z₂ = (-7+7₂).
Express your answer in rectangular form.
m=
Re

Answers

The midpoint m of z₁ = (9+7i) and Z₂ = (-7+7i) is 1 + 7i .

Given complex numbers:

[tex]z_{1}[/tex] = (9 + 7i) and [tex]z_{2}[/tex] = (-7 + 7i)

compare these numbers with a1+ib1 and a2+ib2, we get

a1 = 9, a2 = -7 , b1 = 7 and b2 = 7.

Mid point of complex numbers = a1 + a2 /2 + (b1 + b2 /2)i

= (9 + (-7)/2 + (7 + 7 /2)i

= 2/2 + 14/2 i

Mid point m = 1 + 7i

Therefore the midpoint m of z₁ = (9+7i) and Z₂ = (-7+7i) is 1 + 7i

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write each of the following numbers as a power of the number 2

Answers

Answer

The power on 2 is either -3.5 in decimal form or (-7/2) in fraction form.

Explanation

To do this, we have to first note that

[tex]\begin{gathered} \sqrt[]{2}=2^{\frac{1}{2}} \\ \text{And} \\ 16=2^4 \end{gathered}[/tex]

So, we can then simplify the given expression

[tex]\begin{gathered} \frac{\sqrt[]{2}}{16}=\frac{2^{\frac{1}{2}}}{2^4}=2^{\frac{1}{2}-4} \\ =2^{0.5-4} \\ =2^{-3.5} \\ OR \\ =2^{\frac{-7}{2}} \end{gathered}[/tex]

Hope this Helps!!!

Suppose the coordinate of p=2 and PQ=8. Whare are the possible midpoints for PQ?

Answers

The midpoint for segment PQ can be calculated as:

[tex]\frac{P+Q}{2}[/tex]

Then, the midpoint of PQ is:

[tex]\frac{2\text{ + Q}}{2}=1+0.5Q[/tex]

Additionally, PQ can be calculated as:

[tex]PQ=\left|Q-P\right|[/tex]

So:

[tex]\begin{gathered} \left|Q-P\right|=8 \\ \left|Q-2\right|=8 \end{gathered}[/tex]

It means that:

[tex]\begin{gathered} Q-2=8\text{ or } \\ 2\text{ - Q = 8} \end{gathered}[/tex]

Solving for Q, we get:

Q = 8 + 2 = 10 or Q = 2 - 8 = -6

Finally, replacing these values on the initial equation for the midpoint, we get:

If Q = 10, then:

midpoint = 1 + 0.5(10) = 1 + 5 = 6

If Q = -6, then:

midpoint = 1 + 0.5(-6) = 1 - 3 = -2

The possible midpoints for PQ are 6 and -2

-27\sqrt(3)+3\sqrt(27), reduce the expression

Answers

[tex]-18\sqrt[]{3}[/tex]

Explanation

[tex]-27\sqrt[]{3}+3\sqrt[]{27}[/tex]

Step 1

Let's remember one propertie of the roots

[tex]\sqrt[]{a\cdot b}=\sqrt[]{a}\cdot\sqrt[]{b}[/tex]

hence

[tex]\sqrt[]{27}=\sqrt[]{9\cdot3}=\sqrt[]{9}\cdot\sqrt[]{3}=3\sqrt[]{3}[/tex]

replacing in the expression

[tex]\begin{gathered} -27\sqrt[]{3}+3\sqrt[]{27} \\ -27\sqrt[]{3}+3(3\sqrt[]{3}) \\ -27\sqrt[]{3}+9\sqrt[]{3} \\ (-27+9)\sqrt[]{3} \\ -18\sqrt[]{3} \end{gathered}[/tex]

therefore, the answer is

[tex]-18\sqrt[]{3}[/tex]

I hope this helps you

The cost, c(x) in dollars per hour of running a trolley at an amusement park is modelled by the function [tex]c(x) = 2.1x {}^{2} - 12.7x + 167.4[/tex]Where x is the speed in kilometres per hour. At what approximate speed should the trolley travel to achieve minimum cost? A. About 2km/h B about 3km/h C about 4km/D about 5km/hr

Answers

The equation is modelled by the function,

c(x) = 2.1x^2 - 12.7x + 167.4

The general form of a quadratic equation is expressed as

ax^2 + bx + c

The given function is quadratic and the graph would be a parabola which opens upwards because the value of a is positive

Since x represents the speed, the speed at which the he

I NEED HELP
5C/2 = 20

Answers

you would have to do this backwards

20 times 2 would remove the /2

5c=40

40 divided by 5

is

8

C=8

Can you please solve the last question… number 3! Thanks!

Answers

Let us break the shape into two triangles and solve for the unknowns.

The first triangle is shown below:

We will use the Pythagorean Theorem defined to be:

[tex]\begin{gathered} c^2=a^2+b^2 \\ where\text{ c is the hypotenuse and a and b are the other two sides} \end{gathered}[/tex]

Therefore, we can relate the sides of the triangles as shown below:

[tex]25^2=y^2+16^2[/tex]

Solving, we have:

[tex]\begin{gathered} y^2=25^2-16^2 \\ y^2=625-256 \\ y^2=369 \\ y=\sqrt{369} \\ y=19.2 \end{gathered}[/tex]

Hence, we can have the second triangle to be:

Applying the Pythagorean Theorem, we have:

[tex]22^2=x^2+19.2^2[/tex]

Solving, we have:

[tex]\begin{gathered} 484=x^2+369 \\ x^2=484-369 \\ x^2=115 \\ x=\sqrt{115} \\ x=10.7 \end{gathered}[/tex]

The values of the unknowns are:

[tex]\begin{gathered} x=10.7 \\ y=19.2 \end{gathered}[/tex]

Find the sum of the arithmetic series -1+ 2+5+8+... where n=7.A. 56B. 184C. 92D. 380Reset Selection

Answers

Answer:

Explanation:

The arithmetic series is:

-1 + 2 + 5 + 8 + .....

The first term, a = -1

The common difference, d = 2 - (-1)

d = 3

The number of terms, n = 7

Find the sum of the arithmetic series below

[tex]\begin{gathered} S_n=\frac{n}{2}[2a+(n-1)d] \\ \\ S_7=\frac{7}{2}[2(-1)+(7-1)(3)] \\ \\ S_7=\frac{7}{2}(-2+18) \\ \\ S_7=\frac{7}{2}(16) \\ \\ S_7=56 \end{gathered}[/tex]

Therefore, the sum of the arithmetic series = 56

A square has a perimeterof 8,000 centimeters. Whatis the length of each side ofthe of the square inmeters?

Answers

Answer:

20 meters

Explanation:

The formula for calculating the perimeter of a square is expressed as

perimeter = 4s

where

s is the length of each side of the square

From the information given,

perimeter = 8,000 centimeters

Recall,

1 cm = 0.01 m

8000cm = 8000 x 0.01 = 80 m

Thus,

80 = 4s

s = 80/4

s = 20

The length of each side of the square is 20 meters

I inserted a picture of the question can you please hurry

Answers

Given:

[tex](-2,-5)\text{ and (}1,4)\text{ are given points.}[/tex][tex]\begin{gathered} \text{Slope}=\frac{y_2-y_1}{x_2-x_1} \\ \text{Slope}=\frac{4+5}{1+2} \\ \text{Slope}=\frac{9}{3} \\ \text{Slope}=3 \end{gathered}[/tex]

What is 3 +4.3+45?A4늘OB.B. 7O. 8○ D. 12

Answers

solution

[tex]3+4\frac{1}{3}=7\frac{1}{3}[/tex]

answer: B

Which of the following shows the expansion of sum from n equals 0 to 4 of 2 minus 5 times n ?

(−18) + (−13) + (−8) + (−3) + 0
(−3) + (−8) + (−13) + (−18) + (−23)
2 + (−3) + (−8) + (−13) + (−18)
2 + 7 + 12 + 17 + 22

Answers

The option that indicates the required sum when n equals 0 to 4 of 2 minus 5 times n, is 2 + (−3) + (−8) + (−13) + (−18) (Option C)

What is the Sum of sequences?

The sum of the terms of a sequence is called a series.

From the given sum of a sequence, we are to find the sum of the given sequence from n = 0 to n = 4

When n = 0

a(0) = 2 - 5(0)

a(0) = 2 - 0

a(0) = 2

When n = 1

a(1) = 2 - 5(1)

a(1) = 2 -5

a(1) = -3

When n = 2

a(2) = 2 - 5(2)

a(2) = 2 - 10

a(2) = -8

When n = 3

a(3) = 2 - 5(3)

a(3) = 2 - 15

a(3) = -13

When n = 4

a(4) = 2 - 5(4)

a(4) = 2 - 20

a(4) = -18

Hence the required sum is 2 + (−3) + (−8) + (−13) + (−18)

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The required sum is 2 + (−3) + (−8) + (−13) + (−18)  when n equals 0 to 4 of 2 minus 5 times n, which is the correct answer that would be an option (C).

The given expression is (2 - 5n)

We to determine the sum of the given sequence from n = 0 to n = 4

Let the required sum is  T₀ + T₁  + T₂ +  T₃ + T₄

Substitute the value of n = 0 in the expression (2 - 5n) to get T₀

⇒ T₀ = 2 - 5(0) = 2 - 0 = 2

Substitute the value of n = 1 in the expression (2 - 5n) to get T₁

⇒ T₁ = 2 - 5(1)  = 2 -5 = -3

Substitute the value of n = 2 in the expression (2 - 5n) to get T₂

⇒ T₂ = 2 - 5(2) = 2 - 10 = -8

Substitute the value of n = 3 in the expression (2 - 5n) to get T₃

⇒ T₃ = 2 - 5(3) = 2 - 15 = -13

Substitute the value of n = 4 in the expression (2 - 5n) to get T₄

⇒ T₄ = 2 - 5(4) = 2 - 20 = -18

Therefore, the required sum is 2 + (−3) + (−8) + (−13) + (−18)

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PLEASE HELP AS SOON AS POSSIBLE PLEASE!! ( one question, can whole numbers be classified as integers and rational numbers)

Answers

ANSWER

G. 10 and -2 only

EXPLANATION

-5/4 is a fraction that can't be simplified. Therefore it is not an integer.

1.25 has decimals, so it is not an integer either.

10 and -2 are are integers.

is 2÷2 4 or am I wrong

Answers

2/2 = 1

The answer would be 1

Lne segment AC and BD are parallel, what are the new endpoints of the line segments AC and BD if the parallel lines are reflected across the y-axis?

Answers

Given a point P = (x, y) a reflection P' alongside the y axis of that point follows the rule:

[tex]P=(x,y)\Rightarrow P^{\prime}=(-x,y)[/tex]

We need to multiply the x coordinates of the points by (-1)

The cordinates of the points in the problem are:

A = (2, 5)

B = (2, 4)

C = (-5, 1)

D = (-5, 0)

Then the endpoints of the reflection over the y axis are:

A' = (-2, 5)

B' = (-2, 4)

C' = (5, 1)

D' = (5, 0)

Which is the second option.

Need answer if you could show work would be nice

Answers

(x^3 - 3x^2 - 5x + 39) = (x^2 - 6x + 13)(x + 3). [you can verify this by expanding the brackets]
x + 3 = 0 when x = -3

x^2 - 6x + 13 = 0
x^2 - 6x + 9 + 4 = 0
(x - 3)^2 = -4
This shows x^2 - 6x + 13 = 0 has no real roots
The function only has one real root at x = -3

write the equation for a quadratic function in vertex form that opebs down shifts 8 units to the left and 4 units down .

Answers

STEP - BY STEP EXPLANATION

What to find?

Equation for a quadratic equation.

Given:

Shifts 8 unit to the left.

4 units down

Step 1

Note the following :

• The parent function of a quadratic equation in general form is given by;

[tex]y=x^2[/tex]

• If f(x) shifts q-units left, the f(x) becomes, f(x+q)

,

• If f(x) shift m-units down, then the new function is, f(x) -m

Step 2

Apply the rules to the parent function.

8 units to the left implies q=8

4 units down implies m= 4

[tex]y=(x+8)^2-4[/tex]

ANSWER

y= (x+8)²- 4

Rami practices his saxophone for 5/6 hour on 4 days each week.
How many hours does Rami practice his saxophone each week?

[] 2/[] Hr

Answers

Answer:

you take 5/6 and multiply it by 4/1.

which gives you 20/6

then reduce it by dividing the top number by the bottom number

 which gives you 3 with a remainder of 2

you then place the remainder over the

This tells you he practicedfor 3 2/6

Step-by-step explanation:

I need help with this question... the correct answer choice

Answers

Reflection over the x-axis:

(x,y)--->(x, -y)

and the question is what is not a reflection across the x-axis.

so,

the correct option is D which is:

R'(-9, 4) ----> R'(9, -4)

Because it is a reflection over the y-axis.

Using the distributive property, show how to decompose 8 * 78

Answers

Given any three numbers a, b, and c.

By the distributive law, we must have:

a x (b + c) = (a x b) + (a x c)

Now to find 8 x78

8 x 78 = 8 x (70 + 8) = (8 x 70) + (8 x 8) = 560 + 64 = 624

The slope and y-intercept of the relation represented by the equation 12x-9y+12=0 are:

Answers

12x - 9y +12 =0

To find the slope and y intercept, we want to put the equation in slope intercept form

y = mx+b where m is the slope and b is the y intercept

Solve the equation for y

Add 9y to each side

12x - 9y+9y +12 =0+9y

12x+12 = 9y

Divide each side by 9

12x/9 +12/9 = 9y/9

4/3 x + 4/3 = y

Rewriting

y = 4/3x + 4/3

The slope is 4/3 and the y intercept is 4/3

The answer is 57.3 provided by my teacher, I need help with the work

Answers

Apply the angles sum property in the triangle ABC,

[tex]62+90+\angle ACB=180\Rightarrow\angle ACB=180-152=28^{}[/tex]

Similarly, apply the angles sum property in triangle BCD,

[tex]20+90+\angle BCD=180\Rightarrow\angle BCD=180-110=70[/tex]

From triangle ABC,

[tex]BC=AC\sin 62=30\sin 62\approx26.5[/tex]

From triangle BDC,

[tex]BD=BC\cos 20=26.5\cos 20\approx24.9[/tex]

Now, consider that,

[tex]\angle BDE+\angle BDC=180\Rightarrow\angle BDE+90=180\Rightarrow\angle BDE=90[/tex]

So the triangle BDE is also a right triangle, and the trigonometric ratios are applicable.

Solve for 'x' as,

[tex]x=\tan ^{-1}(\frac{BD}{DE})=\tan ^{-1}(\frac{24.9}{16})=57.2764\approx57.3[/tex]

Thus, the value of the angle 'x' is 57.3 degrees approximately.ang

Find the exact solution to the exponential equation. (No decimal approximation)

Answers

Let's solve the equation:

[tex]\begin{gathered} 54e^{3x+3}=16 \\ e^{3x+3}=\frac{16}{54} \\ e^{3x+3}=\frac{8}{27} \\ \ln e^{3x+3}=\ln (\frac{8}{27}) \\ 3x+3=\ln (\frac{2^3}{3^3}) \\ 3x+3=\ln (\frac{2}{3})^3 \\ 3x+3=3\ln (\frac{2}{3}) \\ 3x=-3+3\ln (\frac{2}{3}) \\ x=-1+\ln (\frac{2}{3}) \\ x=-1+\ln 2-\ln 3 \end{gathered}[/tex]

Therefore the solution of the equation is:

[tex]x=-1+\ln 2-\ln 3[/tex]

ExpenseYearly costor rateGasInsuranceWhat is the cost permile over the course ofa year for a $20,000 carthat depreciates 20%,with costs shown in thetable, and that hasbeen driven for 10,000miles?$425.00$400.00$110,00$100.0020%OilRegistrationDepreciationB. $4.10 per mileA. $1.10 per mileC. $0.25 per mileD. $0.50 per mile

Answers

Step 1:

Find the depreciation

[tex]\begin{gathered} \text{Depreciation = 20\% of \$20,000} \\ \text{Depreciation = }\frac{20}{100}\text{ }\times\text{ \$20000} \\ \text{Depreciation = \$4000} \end{gathered}[/tex]

Step 2:

Total cost = $425 + $400 + $110 + $100 + $4000

Total cost = $5035

Final answer

[tex]\begin{gathered} \text{Cost per mile = }\frac{5035}{10000} \\ \text{Cost per mile = \$0.5035} \\ \text{Cost per mile = \$0.50} \end{gathered}[/tex]

Option D $0.50 per mile

(I don't know if there are tutors here right now at this time but it's worth a try.) Please help me I really really don't understand this, it's going to take me a while to understand this. X(

Answers

[tex]\begin{gathered} 3(b+5)=4(2b-5) \\ 3b+15=8b-20 \\ 15+20=8b-3b \\ 5b=35 \\ b=\frac{35}{5} \\ b=7 \end{gathered}[/tex][tex]3(b+5)=4(2b-5)[/tex]

by the distributive law x(y+z)=zy+xz, we have

[tex]\begin{gathered} 3b+3(5)=4(2b)-4(5) \\ 3b+15=8b-20 \end{gathered}[/tex]

Then we use the properties of inequalities, we can switch both sides, and if we add or multiply something on both sides the equality remains

[tex]\begin{gathered} 3b+15=8b-20 \\ \end{gathered}[/tex]

we want the variables and the numbers without variables to be in different side, so, first we add 20 to both sides, note that the -20 will be cancelled

[tex]\begin{gathered} 3b+15+20\text{ = 8b-20+20} \\ 3b+15+20=8b \end{gathered}[/tex]

we want to left all the numbers with variable on the right side so we substract 3b (add -3b) to both sides. Same as before, the 3b will be cancellated (we can change the order in the sum)

[tex]\begin{gathered} -3b+3b+15+20=-3b+8b \\ 15+20=8b-3b \end{gathered}[/tex]

of course, you're welcome

I was asking if you have understood my explanation so far

tell me

it doesn't matter the order, in fact, when you get used to the method you can work with both at the same time

any other question?

yes, you could substrac 3b first

For example

[tex]\begin{gathered} 2+3x=6-x \\ 2+3x+x=6-x+x \\ 2+3x+x=6 \\ -2+2+3x+x=-2+6 \\ 3x+x=6-2 \\ 4x=4 \\ \end{gathered}[/tex]

sadly I will need to leave since my shift is over, but if you ask another question one of my partners will help you

Have a nice evening!!!!

then we add like terms and switch both sides

[tex]5b=35[/tex]

And then we multiply by 1/5 both sides

[tex]\begin{gathered} 5\frac{1}{5}b=\frac{35}{5} \\ b=\frac{35}{5} \\ b=7 \end{gathered}[/tex]

Other Questions
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