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3 years ago

5

T is the midpoint of rq if r=12 t=-3 what is q

1 answer:

Gemiola [76]3 years ago

5 0

Well how far away is t (-3) from r (12)?
It's 15 away. It gets 15 less, so q will be 15 more.
12+15=27
q is 27

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SVETLANKA909090 [29]

Answer:

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Step-by-step explanation:

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Your company is hosting a benefit 5 kilometer race. How long will the race be to the nearest tenth of a mile?

Akimi4 [234]

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How do I solve this out??

GuDViN [60]

<h3>Answer:</h3>

(x, y) = (7, -5)

<h3>Step-by-step explanation:</h3>

It generally works well to follow directions.

The matrix of coefficients is ...

$\left[\begin{array}{cc}2&4\\-5&3\end{array}\right]$

Its inverse is the transpose of the cofactor matrix, divided by the determinant. That is ...

$\dfrac{1}{26}\left[\begin{array}{ccc}3&-4\\5&2\end{array}\right]$

So the solution is the product of this and the vector of constants [-6, -50]. That product is ...

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... (x, y) = (7, -5)

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In basketball, hang time is the time that both of your feet are off the ground during a jump. The equation for hang time is t =

const2013 [10]

Considering the hang time equation, it is found that Player 1 jumped 0.68 feet higher than Player 2.

<h3>What is the hang time equation?</h3>

The hang-time of the ball for a player of jump h is given by:

$t = 2\left(\frac{2h}{32}\right)^{\frac{1}{2}}$

The expression can be simplified as:

$t = 2\sqrt{\frac{h}{16}}$

For a player that has a hang time of 0.9s, the jump is found as follows:

$0.9 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.9}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.9}{2}\right)^2$

$h = 16\left(\frac{0.9}{2}\right)^2$

h = 3.24 feet.

For a player that has a hang time of 0.8s, the jump is found as follows:

$0.8 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.8}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.8}{2}\right)^2$

$h = 16\left(\frac{0.8}{2}\right)^2$

h = 2.56 feet.

The difference is given by:

3.24 - 2.56 = 0.68 feet.

More can be learned about equations at brainly.com/question/25537936

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2 years ago