All categories

3 years ago

Find the midpoint of the segment with the following endpoints.

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

4 0

Answer:

( 5.5 , 4 )

Step-by-step explanation:

Use mid point formula shown in image

first x= (X+X)/2

x=(4+7)/2

x=11/2

x=5.5

y=(Y+Y)/2

y=(2+6)/2

y=8/2

y=4

(5.5,4)

You might be interested in

Simplified product ?

mel-nik [20]

Answer:

Last choice is correct.

Step-by-step explanation:

$\left(\sqrt{10x^4}-x\sqrt{5x^2}\right)\left(2\sqrt{15x^4}+\sqrt{3x^3}\right)$

$\left(x^2\sqrt{10}-x\cdot x\sqrt{5}\right)\left(2\cdot x^2\sqrt{15}+x\sqrt{3x}\right)$

$\left(x^2\sqrt{10}-x^2\sqrt{5}\right)\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$x^2\sqrt{10}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)-x^2\sqrt{5}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$2x^4\sqrt{150}+x^3\sqrt{30x}-2\sqrt{75}x^4-x^3\sqrt{15x}$

$2x^4\cdot5\sqrt{6}+x^3\sqrt{30x}-2\cdot5\sqrt{3}x^4-x^3\sqrt{15x}$

$10x^4\sqrt{6}+x^3\sqrt{30x}-10\sqrt{3}x^4-x^3\sqrt{15x}$

$10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

Hence final answer is $10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

5 0

3 years ago

Consider a pattern that begins with 180 and each consecutive number is 19 less than the previous term. What are the first five t

Lynna [10]

C is the correct answer. Hope that helped.

4 0

3 years ago

Read 2 more answers

In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the

natka813 [3]

Answer:

$P(t) = 5100e^{0.0039t}$

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that $P(0) = 5100$

$P(t) = 5100e^{rt}$

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that $P(5) = 5200$ .

We use this to find r.

$P(t) = 5100e^{rt}$

$5200 = 5100e^{5r}$

$e^{5r} = \frac{52}{51}$

$\ln{e^{5r}} = \ln{\frac{52}{51}}$

$5r = \ln{\frac{52}{51}}$

$r = \frac{\ln{\frac{52}{51}}}{5}$

$r = 0.0039$

So the equation for the moose population is:

$P(t) = 5100e^{0.0039t}$

5 0

3 years ago

ABCD is a trapezium in which AB parallel DC ,DC=30cm and AB=50cm .If X and Y are respectively the mid point of AD and BC then pr

Ratling [72]

I have attached the answer. Hope it helps.

Download pdf

3 0

3 years ago

What is the sum?<br> 8+(-12)<br> -20<br> 24<br> 4<br> 20

blagie [28]

Answer:

-4

Step-by-step explanation:

adding a negative is the same as subtracting a positive

<h2><u><em>xoxo, </em></u></h2><h2><u><em>your highness...</em></u></h2>

4 0

3 years ago

Read 2 more answers