The average rate of change from x = 1 to x = 3 of f(x)=x^3

How do you multiply -6 with 24.50

GrogVix [38]

The answer to the question

6 0

3 years ago

HELP MI PORFA<3

Maru [420]

Step-by-step explanation:

2(3x+4)=56

<=> 6x+8=56

<=> 6x=48

<=> x=8

3 0

3 years ago

DVDs can be made in a factory in New Mexico at a rate of 20 DVDs for $4.60.

Helen [10]

Answer:

20/4.6 or 4.35

Step-by-step explanation:

unit cost is 20/4.6 which rounds to 4.35

6 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.

Learn more here:

brainly.com/question/9168560

6 0

2 years ago

What is 2/5 ÷ 4/7 equal to

olchik [2.2K]

When you have to divide a fraction, you have to invert the second fraction (4/7 becomes 7/4), and you then multiply the two fractions.

=2/5 ÷ 4/7
=2/5 * 7/4
Multiply the numerators on top
Multiply the denominators on bottom
=(2*7)/(5*4)
=14/20
Simplify by dividing 2 into top & bottom
=7/10

Hope this helps!! :)

3 0

3 years ago

The average rate of change from x = 1 to x = 3 of f(x)=x^3 - 4 is __________