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2 years ago

Write an expression that represents the product of q and 4÷ 5

Mathematics

1 answer:

laiz [17]2 years ago

7 0

Q=20

~~~~Ella~~~~~ maybe I think

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Solve this problem please

Viefleur [7K]

Answer:

x=22

Step-by-step explanation:

1408/64

8 0

3 years ago

Read 2 more answers

Explain why you cannot factor the trinomial x(to the power of 2)−12x+6

Vladimir [108]

Answer:

See below

Step-by-step explanation:

You cannot find two whole number integers that give a sum of -12 and a product of 6.

-1 + -11 = -12, but (-1)(-11) = 11, which isn't 6

-2 + -10 = -12, but (-2)(-10) = 20, which isn't 6

...and so on...

4 0

2 years ago

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BRAINLIEST TO THE BEST ANSWER !!!!

TiliK225 [7]

Which equation am I looking at

4 0

3 years ago

Write an equation for a line passing through (-4, 3) and (0, 6)

Elena L [17]

Answer:

y = 3/4x + 6

Step-by-step explanation:

6-3 3

------- = ------

0--4 4

y = mx + b

take coordinates for y and x out of (-4, 3) or (0,6)

6= 3/4 * 0 + b

6 = 0 + b

b = 6

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

3 years ago