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Fed [463]
3 years ago
11

Find the coordinates of the midpoint of the segment whose endpoints are given.

Mathematics
2 answers:
lidiya [134]3 years ago
7 0

\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ W&({{ -3}}\quad ,&{{ -7}})\quad X&({{ -8}}\quad ,&{{ -4}}) \end{array}\qquad \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \left(\cfrac{-8+(-3)}{2}\quad ,\quad \cfrac{-4+(-7)}{2} \right)\implies \left(\cfrac{-8-3}{2}\quad ,\quad \cfrac{-4-7}{2} \right) \\\\\\ \left( \cfrac{-11}{2}~,~\cfrac{-11}{2} \right)\implies \left( -5\frac{1}{2}~~,~~-5\frac{1}{2} \right)

Lelechka [254]3 years ago
4 0

Answer: (-11/2, -11/2)

Step-by-step explanation: my odyssey class said this was so

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700 tickets to the school play were sold by students. Meryl sold 56 tickets. What percent of the tickets did Meryl sell?
Anni [7]

Answer: 644

Step-by-step explanation:

700 - 56 = 644

4 0
3 years ago
A foam toy is 2 inches wide. It doubles in size for every minute it is in water. Write an expression for the width of the toy af
Solnce55 [7]

it would be 2(2^5), which is two times two to the fifth power. you take the original size and double it for every minute, which is like squaring it, but to the power of five, because it doubles once every minute for five minutes. i hope this makes sense!:)

7 0
4 years ago
HSHILL
AveGali [126]

Answer:

Step-by-step explanation:

4 0
3 years ago
Suppose that an individual has a body fat percentage of 19.3% and weighs 187 pounds. How many pounds of her weight is made up of
Hatshy [7]

Answer:

The individual has 36.09 pounds of body fat.

Step-by-step explanation:

Given that an individual has a body fat percentage of 19.3% and weighs 187 pounds, to determine how many pounds of her weight is made up of fat the following calculation must be performed:

(187 x 19.3) / 100 = X

3.609.1 / 100 = X

36.091 = X

Therefore, the individual has 36.09 pounds of body fat.

6 0
3 years ago
Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
nirvana33 [79]

Answer:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

5 0
4 years ago
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