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3 years ago

Write an equation for “the value of m times the value of n”

Mathematics

1 answer:

NeX [460]3 years ago

7 0

Answer:m(n)

Step-by-step explanation:

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If y varies directly as x, find the constant of variation if y = 25 and x = 5.

Soloha48 [4]

Answer:

The constant of variation is 5

Step-by-step explanation:

We know direct variation is of the form

y = kx

We know y =25 and x=5

25 = k*5

Divide each side by 5

25/5 = 5k/5

5 = k

3 0

3 years ago

Niki makes the same payment every two months to pay off his $61,600 loan. The loan has an interest rate of 9.84%, compounded eve

Alenkasestr [34]

The question is an annuity question with the present value of the annuity given.
The present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt) / (r/t) where PV = $61,600; r = interest rate = 9.84% = 0.0984; t = number of payments in a year = 6; n = number of years = 11 years and P is the periodic payment.
61600 = P(1 - (1 + 0.0984/6)^-(11 x 6)) / (0.0984 / 6)
61600 = P(1 - (1 + 0.0164)^-66) / 0.0164
61600 x 0.0164 = P(1 - (1.0164)^-66)
1010.24 = P(1 - 0.341769) = 0.658231P
P = 1010.24 / 0.658231 = 1534.78
Thus, Niki pays $1,534.78 every two months for eleven years.
The total payment made by Niki = 11 x 6 x 1,534.78 = $101,295.48
Therefore, interest paid by Niki = $101,295.48 - $61,600 = $39,695.48

8 0

3 years ago

Read 2 more answers

What is the value of the expression 8-(2to the 3rd Power+48)divided by 8

dalvyx [7]

8-(8+48) = -48 then divide that by 8 and you get -6 as your result

8 0

2 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$