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3 years ago

5

1 answer:

4 0

This is called the Pythagorean theorem : a ² + b ² = c ². You can substitute any of the variable with any of the known numbers and then you all you have to do is isolate the variable. I hope that helps!!

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Help needed quickly, have to turn in, in 19 minutes

AleksAgata [21]

Answer:

angles F and G are supplementary

Step-by-step explanation:

The angles add up to 180

8 0

3 years ago

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What is the solution of log4x − 616 = 4? x = −2 x = 2 x = 3 x = 4

ivolga24 [154]

This is how to solve for log expressions.
First, do the arithmetic such that all that can be solved is in the other side, besides log.
There are different properties of log to help you solve log expressions.

log4x − 616 = 4
log4x = 620
Change how it is written to solve for x.
x^620 = 4
This makes no sense so we try again.

Or, this could be solved differently.
log 4x-6 16 = 4
(4x-6) ^4 = 16
256x - 1296 = 16
x = 5.125

Final answer: x = 5

3 0

3 years ago

The sum of twice a number and 3 times the same number as 40. What is the number?

Gnom [1K]

Answer:

Let the number be x

According to Question ,

2x + 3x = 40

5x = 40

x = 8

therefore , the number is 8

hope that helps uh...☺

6 0

3 years ago

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Water is being pumped into a conical tank that is 8 feet tall and has a diameter of 10 feet. If the water is being pumped in at

Deffense [45]

The rate of change of the depth of water in the tank when the tank is half

filled can be found using chain rule of differentiation.

When the tank is half filled, the depth of the water is changing at 1.213 ×

10⁻² ft.³/hour.

Reasons:

The given parameter are;

Height of the conical tank, h = 8 feet

Diameter of the conical tank, d = 10 feet

Rate at which water is being pumped into the tank, = 3/5 ft.³/hr.

Required:

The rate at which the depth of the water in the tank is changing when the

tank is half full.

Solution:

The radius of the tank, r = d ÷ 2

∴ r = 10 ft. ÷ 2 = 5 ft.

Using similar triangles, we have;

$\dfrac{r}{h} = \dfrac{5}{8}$

The volume of the tank is therefore;

$V = \mathbf{\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h}$

$r = \dfrac{5}{8} \times h$

Therefore;

$V = \dfrac{1}{3} \cdot \pi \cdot \left( \dfrac{5}{8} \times h\right)^2 \cdot h = \dfrac{25 \cdot h^3 \cdot \pi}{192}$

By chain rule of differentiation, we have;

$\dfrac{dV}{dt} = \mathbf{\dfrac{dV}{dh} \cdot \dfrac{dh}{dt}}$

$\dfrac{dV}{dh}=\dfrac{d}{h} \left( \dfrac{25 \cdot h^3 \cdot \pi}{192} \right) = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64}}$

$\dfrac{dV}{dt} = \dfrac{3}{5} \ ft.^3/hour$

Which gives;

$\dfrac{3}{5} = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64} \times \dfrac{dh}{dt}}$

When the tank is half filled, we have;

$V_{1/2} = \dfrac{1}{2} \times \dfrac{1}{3} \times \pi \times 5^2 \times 8 =\mathbf{ \dfrac{25 \cdot h^3 \cdot \pi}{ 192}}$

Solving gives;

h³ = 256

h = ∛256

$\dfrac{3}{5} \times \dfrac{64}{25 \cdot h^2 \cdot \pi} = \dfrac{dh}{dt}$

Which gives;

$\dfrac{dh}{dt} = \dfrac{3}{5} \times \dfrac{64}{25 \cdot (\sqrt[3]{256}) ^2 \cdot \pi} \approx \mathbf{1.213\times 10^{-2}}$

When the tank is half filled, the depth of the water is changing at 1.213 × 10⁻² ft.³/hour.

Learn more here:

brainly.com/question/9168560

6 0

2 years ago

I need help bad very bad

Nat2105 [25]

I think it's the second one

6 0

3 years ago

Read 2 more answers