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2 years ago

Which statements about the formulas for finding area are true? Select two options.

Mathematics

2 answers:

omeli [17]2 years ago

8 0

Answer:

The area formulas for a triangle and a parallelogram require multiplying base by height.

A rectangle and parallelogram with the same base and height will have the same area.

lara31 [8.8K]2 years ago

4 0

Answer:

a and c to make it easier

Step-by-step explanation:

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Is 5/7 greater than 10/13

NeTakaya

Answer:

No is false

Step-by-step explanation:

5 / 7 > 10 / 13 = False

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2 years ago

Write the expression as a product of two factors. 2* 1 + (1+7) + (762)

Arlecino [84]

Answer:

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Step-by-step explanation:

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2 years ago

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What is the answer to the equation x squared minus 2x equals 15.

anzhelika [568]

Answer:

$x = 5 \\ x = - 3$

Step-by-step explanation:

${x}^{2} - 2x = 15 \\ {x}^{2} - 2x - 15 = 0 \\ {x}^{2} + 3x - 5x - 15 = 0 \\ x(x + 3) - 5(x + 3) = 0 \\ (x - 5)(x + 3) = 0 \\ x - 5 = 0 \: \: \: \: \: or \: \: \: \: x + 3 = 0 \\ x = 5 \: \: \: \: or \: \: \: \: \: \: x = - 3$

7 0

2 years ago

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.

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2 years ago

Henry subracts 16 from it and gets an answer of 13.9 what was the orginal number

sergiy2304 [10]

Answer:

29.9 is the answer for that question

5 0

2 years ago

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