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2 years ago

7

(HELP PLS WILL MARK BRAINLIEST) For her birthday, Allie received $75. She wants to buy some CDs

2 answers:

IrinaK [193]2 years ago

7 0

Answer: B

Step-by-step explanation:

Times $74.95 by each number, and the one closet will be your answer:

$14.95 x 5 = $74.95 rounded to the nearest whole dollar is $75

Yuki888 [10]2 years ago

3 0

Answer:

5

Step-by-step explanation:

She received $75

1 CD cost $14.95

Therefore the number of CDs that can be bought are $75/$14.95 is

5.016

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The perimeter of the rectangle below is 46 units. Find the length of side RS. Write your answer without variables.

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Step-by-step explanation:

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3 years ago

Please I need help on solving for x

Sedaia [141]

Answer:

$34^o=1/2((3x-2)^0-(x+6)^o)$

$68^o=3x^o-2^o-x^o-6^o$

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$\huge\boxed{\boxed{\underline{\textsf{\textbf{x=38}}}}}$

<h3><u>---------------------</u></h3><h3><u>hope it helps...</u></h3><h3><u>have a great day!!</u></h3>

5 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:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

<u />

<u>(a) The books in any order</u>

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:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

<u>(b) The mathematics book, not together</u>

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

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

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

<u>(c) The novels must be together and the chemistry books, together</u>

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:

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

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

$\mathbf{Arrangements = 17280}$

<u>(d) The mathematics must be together and the chemistry books, not together</u>

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:

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

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

$\mathbf{Total = 302400}$

Read more about permutations at:

brainly.com/question/1216161

8 0

1 year ago

I need to show how to find these solutions 1/4 and 3/4 by completing the square!

Arlecino [84]

Answer:

See explanation, and ask for more details if unclear!

Step-by-step explanation:

The perfect square of this equation is $x^2-x+\frac{1}{4}$ , since the square would be $(x-\frac{1}{2})^2$ . 1/4=4/16, meaning that you can set up the equation in the following way:

$(x^2-x+\frac{1}{4})-\frac{1}{16}=0$

$(x-\frac{1}{2})^2=\frac{1}{16}$

Take the square root of both sides:

$x-\frac{1}{2}=\pm \frac{1}{4}$

Add 1/2 to both sides:

$x=\frac{1}{2}\pm \frac{1}{4}=\frac{3}{4}, \frac{1}{4}$ . Hope this helps!

8 0

2 years ago

Read 2 more answers