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2 years ago

Select all of the equations for which x = 7 is a solution?

Mathematics

2 answers:

9966 [12]2 years ago

8 0

Answer:

d x – 3 = 4

e 4x + 1 = 29

hope it helps

Alina [70]2 years ago

7 0

Answer:

Step-by-step explanation:

x - 3 = 4

reciprocate the 3 it will be +3, so the solution will be

x= 4+3

4+3=7

therefore the equation 4+3=7 states that the x is equal to 7

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Brainliest to first correct answer.

tester [92]

The answer is x = -0.6 and y = -8.2.

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3 years ago

How are expressions 1/4 of 12 and 12 divided by 4

Radda [10]

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3 years ago

P = q/4 -r solve for q

Ivan

Answer:

4(p+r) = q

Step-by-step explanation:

p = q/4 -r

Add r to each side

p+r = q/4 -r+r

p+r = q/4

Multiply each side by 4

4(p+r) = q/4*4

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3 years ago

Ivana and Gloria are sisters. Gloria is

Marina CMI [18]

Answer:

11 years old

Step-by-step explanation:

3 years ago, Gloria was 32 years old

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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

3 years ago