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Stells [14]
3 years ago
6

Simplify completely*****​

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0

Answer:

\displaystyle \frac{2(x + 4)}{3}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Factoring

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle \frac{4x + 16}{6}<u />

<u />

<u>Step 2: Simplify</u>

  1. [Fraction] Factor numerator:                                                                           \displaystyle \frac{4(x + 4)}{6}
  2. [Fraction] Reduce:                                                                                          \displaystyle \frac{2(x + 4)}{3}
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The perimeter of △MTV is 605 in.<br><br> MV = ?<br><br> m∠M = ?
swat32

Answer:

MV = 160 (and %50 sure) X = 27

Step-by-step explanation:

To find MV we just have to subtract the two lengths of the perimeter that we do know, minus the total perimeter, and find the difference. (using that its, 605-160-285 = 160) with that knowledge, we can say two things. MV = 160, and <V = <T by the two side lengths being congruent. With that, we can say that X + X +X+99 = 180, and then just find X, 27+27+27+99=180.

8 0
4 years ago
Find the value of x that will make A||B
olya-2409 [2.1K]

Answer:

7

Step-by-step explanation:

Alternate interior angles must be congruent.

3x - 2 = 2x + 5

x = 7

5 0
3 years ago
If f(7) = 22, then<br> f-1(22) = [?]
NikAS [45]

Answer:its 56

Step-by-step explanation:

8 0
3 years ago
In a box there are six envelopes each containing two cards. Three of the envelopes contain two red cards, two of them contain a
Sphinxa [80]

Answer:

\frac{2}{5}

Step-by-step explanation:

3 envelopes having 2 red card

2 envelopes having 1 red card and 1 black card

1 envelope having 2 black cards

We are given that . An envelope is selected at random and a card is withdrawn and found to be red.

So, No. of ways of envelope having red card = 3+2 = 5

No. of required ways of envelope having 1 red card and 1 black card = 2

So, probability of getting an envelope having 1 red card and 1 black card = \frac{2}{5}

Hence The chance the other card is black is \frac{2}{5}

5 0
3 years ago
Can someone please help me with this question
masya89 [10]

Answer:

I’d say C

Step-by-step explanation:

You should divide 112340 by 125. Then multiply it by 157. The result will be the closest to C.

5 0
3 years ago
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