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3 years ago

Out of 20 cookies, 4/ 5 are chocolate chip. How many cookies are chocolate chip?

Mathematics

2 answers:

Zinaida [17]3 years ago

3 0

Answer:

16 cookies. Try your best.

liq [111]3 years ago

3 0

Answer: 16

Step-by-step explanation:

We can solve this by using the equation: 4/5 x 20 = y

(y is the answer)

Now we can solve it. You can just solve 4/5 x 20, or if you like using decimals you can change it to 0.8 x 20, becaues 4/5 is equivalent to 0.8.

0.8 x 20 = 16.

Therefore, out of the 20 cookies, 16 are chocolate chip.

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Answer:

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A sweater was discounted by 36%. The amount if the discount was $27. What was the cost of the sweater?

neonofarm [45]

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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' ≈ 9.34 × 10⁻⁵ m/s

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

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

<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' ≈ 9.34 × 10⁻⁵ m/s

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;

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

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

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

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

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

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' ≈ 1.97 × 10⁻³ m/s

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 0.175 m

Learn more about the rules of differentiation here:

brainly.com/question/20433457

brainly.com/question/13502804

3 0

3 years ago

If x = 7 − p and y = 3p + 4, which of the following

Rudiy27

Answer:

y = 25 - 3x

Step-by-step explanation:

x = 7 - p ⇒ p = 7 - x

y = 3(7-x) + 4

y = 21 - 3x + 4

y = 25 - 3x

7 0

3 years ago