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3 years ago

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Esther gets her favorite ice cream in a cone. The cone is filled to the top and has a radius of 6.5 cm and a height of 11.25 cm.

What is the volume of the ice-cream?

1 answer:

Yanka [14]3 years ago

6 0

Answer:

73.125

Step-by-step explanation:

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Can someone help me solve this??

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50

Step-by-step explanation:

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A radar gun was used to record the speed of a runner (in meters per second) during selected times in the first 2 seconds of a ra

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3 years ago

Ruben bought 6 new books for his collection. This increased his collection by 12%. How many books does he have now?

Helen [10]

Let's create a variable to represent the size of Ruben's book collection prior to his buying the new books.

B = number of books in Ruben's collection before buying new ones

So, we know that when he bought 6 books, this increased his collection by 12%.

12% of his book collection would be represented by 0.12B. Therefore, we know that:

0.12B = 6

Divide both sides by 0.12, to get to B = 50.

However, you are being asked how many books does he have now. B represents the number of books he had before. Therefore you want to add back the 6 books he just bought.

The answer is 50 + 6 = 56

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3 years ago

20 points<br> For the function f defined by f(x)=3x2−2x+5 find f(−x),−f(x) , and −f(−x).

vekshin1

F (x) -f(x) add the negative on the outside of the F

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3 years ago

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If the price is increasing at a rate of 2 dollars per month when the price is 10 dollars, find the rate of change of the demand.

densk [106]

Answer:

The demand reduces by $7.12 per month

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Step-by-step explanation:

Given

$p\to price$

$x \to demand$

$2x^2+5xp+50p^2=24800.$

$p =10; \frac{dp}{dt} = 2$

Required

Determine the rate of change of demand

We have:

$2x^2+5xp+50p^2=24800.$

Differentiate with respect to time

$4x\frac{dx}{dt} + 5x\frac{dp}{dt} + 5p\frac{dx}{dt} + 100p\frac{dp}{dt} = 0$

Collect like terms

$4x\frac{dx}{dt} + 5p\frac{dx}{dt} = -5x\frac{dp}{dt} - 100p\frac{dp}{dt}$

Factorize

$\frac{dx}{dt}(4x + 5p) = -5(x + 20p)\frac{dp}{dt}$

Solve for dx/dt

$\frac{dx}{dt} = -\frac{5(x + 20p)}{4x + 5p}\cdot \frac{dp}{dt}$

Given that: $2x^2+5xp+50p^2=24800.$ and $p = 10$

Solve for x

$2x^2 + 5x * 10 + 50 * 10^2 = 24800$

$2x^2 + 50x + 5000 = 24800$

Equate to 0

$2x^2 + 50x + 5000 - 24800 =0$

$2x^2 + 50x -19800 =0$

Using a quadratic calculator, we have:

$x \approx -113\ and\ x\approx88$

Demand must be greater than 0;

So: $x=88$

So, we have: $x=88$ ; $p =10; \frac{dp}{dt} = 2$

The rate of change of demand is:

$\frac{dx}{dt} = -\frac{5(88 + 20*10)}{4*88 + 5*10} * 2$

$\frac{dx}{dt} = -\frac{5(288)}{402} * 2$

$\frac{dx}{dt} = -\frac{2880}{402}$

$\frac{dx}{dt} \approx -7.16$

<em>This implies that the demand reduces by $7.12 per month</em>

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3 years ago