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3 years ago

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A rectangular prism has a volume of 150 cm3. if the height measures 3 cm and the width measures 2 cm, what must the length be to

achieve the given volume?

1 answer:

ss7ja [257]3 years ago

3 0

A Legnth of 25 must be achieved

3 x 2 x 25 = 150

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Mr. and mrs. lorenzo want to buy a home valued at $213,500. if they have 18% of this amount saved for a down payment, how much h

valentinak56 [21]

Mr. and Mrs. Lorenzo have saved $38,430 for the home they want to buy

Information about the problem:

Initial cost = $213,500
Savings = 18%.
Final savings =?

To solve this exercise we have to apply the percentage formula:

Final savings = initial cost * savings / 100

Final savings = $213,500 * 18 / 100

Final savings = $213,500 * 0.18

Final savings = $38,430

<h3>What is a percentage?</h3>

Percentage is defined as the number that represents a ratio of a total that is divided by 100 equal parts. It is represented by the symbol %.

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2 years ago

In the equation below a is a number between 0 and 1 8 x a = b which statement about the value of b is always true

Kamila [148]

Answer:

See Explanation

Step-by-step explanation:

Given

$0 \le a \le 1$

$8*a =b$

Required

Determine the true statement about b

The question is incomplete as options are not given. However, a general explanation is as follows:

Rewrite $8*a =b$

$b = 8 * a$

When a = 0

$b = 8 * 0 = 0$

When a = 1

$b = 8 * 1 = 8$

This means that:

For $0 \le a \le 1$

Then $0 \le b \le 8$

<em>This means that: b is a number between 0 and 8</em>

3 0

3 years ago

Given P(x) = 2x3+ 3x2- x + 4, find P(0) and P(-2).

pshichka [43]

9514 1404 393

Answer:

P(0) = 4

P(-2) = 2

Step-by-step explanation:

Put the value where the variable is and do the arithmetic.

P(0) = 2·0³ +3·0² -0 +4

P(0) = 4

__

P(-2) = 2·(-2)³ +3·(-2)² -(-2) +4 = 2(-8) +3(4) +2 +4

P(-2) = -16 +12 +2 +4

P(-2) = 2

6 0

3 years ago

A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t

skad [1K]

The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m/s</u>

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

$1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h$

From the shape of the tank, we have;

$\dfrac{h}{r} = \dfrac{1.5}{1}$

Which gives;

h = 1.5·r

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}$

$\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}$

$\dfrac{dV}{dt} = 0.00011$

Which gives;

$\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}$

When r = 0.5 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}$

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) When the height is 20 cm, we have;

h = 1.5·r

$r = \dfrac{h}{1.5}$

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}$

r = 20 cm ÷ 1.5 = $13.\overline3$ cm = $0.1\overline3$ m

Which gives;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}$

$\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }$ <em />

$\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}$

When the height is 20 cm = 0.2 m, we have;

$\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}$

The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

$\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}$

When the radius is 10 cm = 0.1 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}$

The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m</u>

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3 0

3 years ago

Which expressions are equal to the expression –16.8 – 4.7? Choose all answers that are correct.

ad-work [718]

$Use\ commutative\ property:a+b=b+a\\\\-16.8-4.7=-16.8+(-4.7)\\\\therefore:\\\\a.\ -4.7-16.8=-4.7+(-16.8)=-16.8+(-4.7)=-16.8-4.7\ CORRECT\\\\b.\ -4.7+16.8\ NOT\\\\c.\ -16.8+(-4.7)=-16.8-4.7\ CORRECT\\\\d.\ 16.8-(-14.7)=16.8+14.7\ NOT$

$Your\ answer:\huge\boxed{a\&c}$

7 0

3 years ago