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2 years ago

Find the indicated root

Mathematics

2 answers:

Tanzania [10]2 years ago

8 0

Answer:

${n}^{25}$

Step-by-step explanation:

$\sqrt[3]{ {n}^{75} } \\ {n}^{ \frac{75}{3} } \\ = {n}^{25}$

hope this helps

brainliest appreciated

good luck! have a nice day!

Sever21 [200]2 years ago

6 0

So you have to put 25 inside the blank box.

look at the attached picture

Hope it will help you

Good luck on your assignment

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Can’t answer this as it gives no information?

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2 years ago

Después de 15 meses de haber prestado un capital al 5% de rédito, tengo que pagar de interés Q468.75. ¿Qué capital se ha prestad

Aleonysh [2.5K]

The capital borrowed, or the principal when the interest was Q468.75 is <u>Q7500</u>.

The principal P, amount borrowed, at a certain rate of interest R%, for a time of T years, giving an interest of I, can be calculated using the formula:

P = (I * 100)/(R * T).

In the question, we are asked to find the capital borrowed, that is, the principal, when the user pays Q468.75 after 15 months at 5% of income.

Thus, Interest (I) = Q468.75, rate of interest (R) = 5%, and time (T) = 15 months = 15/12 years = 1.25 years.

Thus, the principal P, can be calculated by substituting the values in the formula: P = (I * 100)/(R * T).

P = (468.75*100)/(5*1.25),

or, P = 46875/6.25,

or, P = 7500.

Thus, the capital borrowed, or the principal when the interest was Q468.75 is <u>Q7500</u>.

The provided question is in Spanish. The question in English is:

"After 15 months of having borrowed capital at 5% of income, I have to pay Q468.75 in interest. What capital has been borrowed?"

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1 year ago

Determine the zeros of the function f(x)=x^2-2x-8.

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2 years ago

A spherical balloon is being inflated at a rate of 3 cubic inches per second. Determine the change in the rate of the radius.How

Novay_Z [31]

Answer:

The rate rate of change of radius is $\frac{1}{48\pi}$ inches per second when the diameter is 12 inches.

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Step-by-step explanation:

Consider the provided information.

A spherical balloon is being inflated at a rate of 3 cubic inches per second.

The volume of sphere is $V=\frac{4}{3}\pi r^3$

Differentiate the above formula with respect to time.

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Substitute the respective values in the above formula,

$3=4\pi 6^2\frac{dr}{dt}$

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The rate rate of change of radius is $\frac{1}{48\pi}$ inches per second when the diameter is 12 inches.

When d=16

$3=4\pi 8^2\frac{dr}{dt}$

$\frac{3}{256\pi}=\frac{dr}{dt}$

Thus, the radius is changing more rapidly when the diameter is 12 inches.

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3 years ago

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