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3 years ago

224 divided by 7 in distributive property

Mathematics

1 answer:

zhenek [66]3 years ago

8 0

(56/7)+(168/7)=8+24=32. Your answer is: 32

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Twice the complement of angles A is 40 less than the supplement of angle A. Find the measure of angle A

Serggg [28]

Complement adds to 90
complement of A is 90-A

supplement is adds to 180
supplemetn of A is 180-A

2(90-A) is -40+180-A

2(90-A)=-40+180-A
2(90-A)=140-A
distribute
180-2A=140-A
add 2A to both sides
180=140+A
minus 140 both sides
40=A

A=40 degres

4 0

3 years ago

Read 2 more answers

A spring stretches in relation to the weight hanging from it all cording to the equation Y equals 0.75X +0.25 where X is the wei

Tomtit [17]

What is the question?

3 0

2 years ago

A Find the slope of the graph using the

Elodia [21]

Answer: 2

Step-by-step explanation:

To find the slope, we use the formula $m=\frac{y_2-y_1}{x_2-x_1}$ . With the given points, we can directly plug them in and solve.

$m=\frac{10-2}{5-1} =\frac{8}{4}=2$

Now, we know that the slope is 2.

7 0

3 years ago

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

2 years ago

Please help me quickly first person to answer gets branliest

ivolga24 [154]

The answer should be 117.6. tell the teacher that the amount of meters is wrong or the answers are wrong

sorry :(

3 0

3 years ago