35/31 equals X over 12

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with

$\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4$

3 0

4 years ago

What times to 18 but adds to 19

Bingel [31]

1 and 18 1 × 18 = 8 1+ 18 = 19

8 0

3 years ago

Please can you help me with this because I had to post twice

steposvetlana [31]

Answer:

y=11

Step-by-step explanation:

3y=y+22

3y-y=y-y+22

2y=22

y=11

4 0

3 years ago

Tina found $4.75 at the bottom of her purse and added it to her wallet. She later bought milk and eggs for $6,55 and had

wariber [46]

Step-by-step explanation:

step 1. x + 4.75 - 6.55 = 5.43

step 2. looks like b is the only answer.

7 0

4 years ago

Read 2 more answers

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D.

yKpoI14uk [10]

The correct statement is:

There are four ways to choose the committee.
There are three ways to form the committee if person D must be on it.
If persons B and C must be on the committee, there are two ways to form the committee.

It is given that:

A membership committee of three is formed from four eligible members.

1) There are four ways to choose the committee.

This statement is true.

Since we have to choose 3 members out of the 4 members so we can use the method of combination.

2) There are three ways to form the committee if person D must be on it.

This statement is also true.

Since D has to be in the committee, this means we have to choose 2 more people out of the three people to form the committee.

3) If seven members are eligible next year, then there will be fewer combinations.

This statement is wrong.

Since we have to choose 3 members out of 7 members so the number of possible combinations will be:

There are 35 combinations possible.

4) If persons B and C must be on the committee, there are two ways to form the committee.

If B and C have to be in the committee then we have to choose just one person out of the two people left.

Hence, the statement is true.

5) If persons A and C must be on the committee, then there is only one way to form the committee.

If A and C have to be on the committee then as in the last option we have to choose any one of the two-person left.

So possible number of ways is 2.

Hence, the statement is false.

Learn more about committee here: brainly.com/question/425830

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<em>Your question is incomplete. Please read below to find the missing content.</em>

<em />

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D. The possible outcomes include S = {ABC, ABD, ACD, BCD}.

Which statements about the situation are true? Check all that apply.

There are four ways to choose the committee.

There are three ways to form the committee if person D must be on it.

If seven members are eligible next year, then there will be fewer combinations.

If persons B and C must be on the committee, there are two ways to form the committee.

If persons A and C must be on the committee, then there is only one way to form the committee.

8 0

2 years ago