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3 years ago

What is the solution to 7 • p = –56? a. p = –49 b. p = –8 c. p = 8 d. p = 49

Mathematics

1 answer:

Lostsunrise [7]3 years ago

8 0

The correct answer is p=-8

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The actual distance between the two houses is 44.1ft. The contractor measures the distance as 46ft. Find the relative error of t

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The correct answer is: 44.1ft - 46ft = -1.9ft/44.1ft = .043% relative error. Please mark Brainliest if helpful

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2 years ago

Hadley paddled a canoe 2/3 mile in 1/4 hour. How fast did Hadley paddle, in miles per hour?

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Answer:

Step-by-step explanation:

Hadley paddled a canoe 2/3 mile in 1/4 hour. it will be 4 times more in an hour because 1/4*4=4/4=1. 2/3*4=8/3=2 2/3

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3 years ago

What is the perimeter of a polygon with vertices at (3, 2) , (3, 9) , (7, 12) , (11, 9) , and (11, 2) ?

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the answer is 32 units.

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3 years ago

( 1 0 3) + (4 1 7) The first entry of the resulting matrix is:

Mamont248 [21]

Hello!

When adding matrices you add the numbers that are in the same position

1 + 4 = 5

0 + 1 = 1

3 + 7 = 10

The answer is (5 1 10)

Hope this helps!

7 0

3 years ago

Read 2 more answers

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago