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2 years ago

Please hurry on an exam The formula for calculating simple interest is I = prt. Which is the equivalent equation solved for r?

Mathematics

2 answers:

ehidna [41]2 years ago

6 0

Answer:

r = C/2π.

Step-by-step explanation:

Brut [27]2 years ago

3 0

Answer:

r=I/pt

Step-by-step explanation:

I = PRT is the formula for interest, and we need to solve it for R. Since all three of P, R, and T are multiplied, we send things to the other side of the equals through dividing. I = PRT I / P = RT to divide both sides by P I / PT = R to divide both sides by T So I / PT = R (the first choice) is the equation solved for R.

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Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.$

3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$

Answer: correct option is B.

7 0

3 years ago

Yan is climbing down a ladder. Each time he descends 4 rungs on the ladder, he stops to see how much farther he

almond37 [142]

Answer:

A. 4+4+4+4 ...............

4 0

3 years ago

Read 2 more answers

What is the multiplier?<br><br>A. 0.123<br><br>B. 0.111<br><br>C. 9<br><br>D. 81

amm1812

Answer:

it is eather d or b

Step-by-step explanation:

7 0

3 years ago

Please help me with this maths question

Keith_Richards [23]

Answer:

add all of them together then divide bye the number of lines

Step-by-step explanation:

8 0

3 years ago

Does anyone know how to simplify this expression? I know it has to do with properties of summation and special summation rules b

dusya [7]

<h3>Answer: -n/(n+1)</h3>

============================================================

Explanation:

Let's expand this out up to say n = 6 terms

I'll write the summation in such a way that each (1/(k+1) - 1/k) grouping will get its own row. Each row has the same value of k.

If we plug in k = 1 through k = 6, but not evaluate just yet, we will have this:

1/(1+1) - 1/1

1/(2+1) - 1/2

1/(3+1) - 1/3

1/(4+1) - 1/4

1/(5+1) - 1/5

1/(6+1) - 1/6

All I've done so far is replace k with 1 through 6. Again, each row represents a different k value. Each row has the general format 1/(k+1)-1/k.

Let's simplify everything in the first column.

So the (1+1) turns into 2, the (2+1) turns into 3, and so on.

Doing that leads to...

1/2 - 1/1

1/3 - 1/2

1/4 - 1/3

1/5 - 1/4

1/6 - 1/5

1/7 - 1/6

Then note how we have the cancellations shown in the diagram below. The color coding shows how the terms pair up to cancel out. By "cancel out", I mean specifically that the fractions add up to 0. Eg: 1/2 + (-1/2) = 0.

Nearly everything cancels out. The only things left after the dust settles is the -1/1 in the first row and the 1/7 in the bottom row. This evaluates to 1/7-1/1 = 1/7 - 7/7 = (1-7)/1 = -6/7

If you were to try this with n = 7, then you should find that again nearly everything cancels but the fractions -1/1 and 1/8. Then 1/8 - 1/1 = -7/8.

For n = 8, you should end up with 1/9 - 1/1 = -8/9.

You can probably see the pattern.

A conjecture is that the answer is -n/(n+1). The general proof of this isn't too tricky. Simply follow the same ideas as mentioned above. Expand out a few terms and see how things cancel out. You'll find that everything cancels except for the -1/1 in the first row and the 1/(n+1) just one term before the very end. Then we can say:

1/(n+1) - 1/1

1/(n+1) - (n+1)/(n+1)

(1-(n+1))/(n+1)

(1-n-1)/(n+1)

-n/(n+1)

If we tried n = 6, then we find,

-n/(n+1) = -6/(6+1) = -6/7

which was the result we found earlier when we added the first n = 6 terms of this series. Trying out n = 7 should lead to -n/(n+1) = -7/8, and so on.

3 0

3 years ago