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1 year ago

A change purse contains a total of 35 nickels and dimes. The total value of the coins is $2.25

Mathematics

1 answer:

Montano1993 [528]1 year ago

6 0

Answer: 25 nickels

10 dimes

Step-by-step explanation:

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You invest $12,000 at 3.8% annual interest compounded annually. How much will you have at the end of 5 years?

Serjik [45]

Answer:

The answer is 60,000.

Step-by-step explanation:

If you invest 12,000 at 3.8 annual interest. Annual means yearly, so that's every year. 12,000 x 5 is 60,000 that's why the answer is 60,000. If the answer is incorrect or correct let me know. If it's correct, mark me as brainlest, give me a rating, or say thank you! Have A Great Day.

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2 years ago

The table shows all the possible outcomes for spinning the pointer of the spinner with four equal sized sections labeled 1–4 and

vladimir1956 [14]

Answer:

12.5%

Step-by-step explanation:

There are two types of spinners here and the outcome of them is independent. That means

P(odd number,C) = P(odd number) * P(C)

There are two odd numbers out of four numbers in the first spinner. The chance of odd number will be:

P(odd) = 2/(2+2)= 1/2= 50%

There are four letters and the desired outcome is C. The chance for C will be:

P(C)= 1/4= 25%

Then the chance will be:

P(odd number,C) = P(odd number) * P(C)

P(odd number,C) =50% * 25% = 12.5%

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2 years ago

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If x-1/x=5, find the value of x^3-1/x^3

Alexxx [7]

65 is the answer. Solve the first equation and then plug in.

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2 years ago

Solve the following by using mathematical Induction. For >/ 1

cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

in left part $1^2=1;$

in right part $\dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.$

2. Assume that for all $k$ following equality is true

$1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

3. Prove that for $k+1$ the following equality is true too.

$1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$

Consider left part:

$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$