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3 years ago

Geometry exam help !!!!please

Mathematics

1 answer:

natta225 [31]3 years ago

4 0

Answer:

you add 17+6 and it would be x=23

Step-by-step explanation:

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Mrs anglin is driving her car at a constant speed of 90 miles in 2 hours select three statements that are true based on her spee

Ira Lisetskai [31]

Answer:

She traveled 45 miles in 1 hour
She traveled 270 miles in 6 hours
She traveled 360 miles in 8 hours

Step-by-step explanation:

If Mrs. Anglin traveled at a speed of 90 miles in 2 hours, it means that her speed per hour is:

= 90/2

= 45 mph

If she is travelling at 45mph then in 1 hoour she would have traveled 45 miles.

In 6 hours she would have traveled:

= 45 * 6

= 270 miles

In 8 hours she would have traveled:

= 45 * 8

= 360 miles

6 0

3 years ago

Lamar got a 75, 89, 94, 95, 88, 96, 94, 82, 98, and 96 on his science tests last quarter. What is the mean absolute deviation of

Alik [6]

The answer to this is 5 ( i think )

4 0

2 years ago

Read 2 more answers

Thanks for the help!

Bogdan [553]

Hope you could understand.

If you have any query, feel free to ask.

5 0

3 years ago

A movie started at 22 45. The movie ended at 01 32 the next day. Calculate the duration of the movie.

elena-s [515]

Answer:

2 hours and 47 minutes.

Step-by-step explanation:

If it makes it easier then translate it to normal time.

10:45 pm and 1:32 am.

2 hrs and 47 minutes.

6 0

3 years ago

Read 2 more answers

The demand function for q units of a product at $p per unit is given by p(q + 3)2 = 100,000. Find the rate of change of quantity

wariber [46]

Answer:

-0.625 unit

Step-by-step explanation:

Given that:

$p(q+3)^2 = 100, 000$

where;

p = price

q = quantity

To find the rate of change of quantity (q) with respect to price (p) we go by the differentiation

$(q+3)^2\frac{dp}{dp} +p^2(q+3)\frac{dq}{dp}=0$

$(q+3)^2 +p^2(q+3)\frac{dq}{dp}=0$

$\frac{dq}{dp}=\frac{-(q+3)}{2p}$

when P =40

Then 4(q+3)² = 100, 000

(q+3)² = $\frac{100,000}{4}$

(q+3)² = 25,000

(q+3) = $\sqrt{2500}$

q+ 3 = 50

q = 50 -3

q = 47

NOW; $\frac{dq}{dp}=\frac{-(q+3)}{2p}$

$\frac{dq}{dp}=\frac{-(47+3)}{2*40}$

$\frac{dq}{dp}=\frac{-47-3}{2*40}$

$\frac{dq}{dp}=\frac{-50}{80}$

$\frac{dq}{dp}=-0.625$

Thus, the rate of change of quantity with respect to price when p = $40 is -0.625 unit.

8 0

4 years ago