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Brilliant_brown [7]
3 years ago
11

Pls I need help I don’t understand this

Mathematics
1 answer:
mr Goodwill [35]3 years ago
7 0
Here’s how to solve:
3x-15+-9
3x-24




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If x = 7 and y = 10, what is the value of the expression xy - ( y + 2) ?
BartSMP [9]

X=7 and y=10

Given that xy-(y+2)

Therefore,

7*10-(10+2)

70-12

58 Ans


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4 years ago
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Wilbur spends 2/3 of his income, share 1/12, and saves the rest. What part of his income does he save? Give the answer in simple
scoundrel [369]

Answer:

<em><u>1/4 of his income.</u></em>

Step-by-step explanation:

If Wilbur spends 2/3 of his income, 1/3 or 4/12 of it is left for other purposes (It is easier if everything has a common denominator of 12). And if he shares 1/12 of that remaining amount, there is 3/12 left. And when we simplify 3/12, we get <u>1/4</u>.

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3 years ago
(1/27)^x-6=27<br><br>Could I please have a step by step as well? Much appreciated! ​
Andrei [34K]

Answer:

x = 7

Step-by-step explanation:

1) Use Division Distributive Property: (x/y)^a = x^a/y^a.

\frac{1}{ {27}^{x - 8} }  = 27

2) Multiply both sides by 27^x - 8.

1 =  {27} \times 27^{1 + x  - 8}

3) Use the product rule: x^a x^b = x^a+b.

1 =  {27}^{1 + x - 8}

4) Simplify 1 + x - 8 to x - 7.

1  =  {27}^{x - 7}

5) Use Definition of Common Logarithm: b^a = x if and only if log<u>b</u><u> </u>(x) = a.

log_{27}1 = x - 7

6) Use Change of Base Rule.

\frac{ log_1  }{  log{27} }  = x - 7

7) Use rule of 1: log 1 = 0.

\frac{0}{ log_{27}}  = x - 7

8) Simplify 0/log_27 to 0.

0 = x - 7

9) Add 7 to both sides.

7 = x

10) Switch sides.

x = 7

<u>Therefor</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>7</u><u>.</u>

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3 years ago
ANSWER TO BE NAMED BRAINLIEST GET A THANK YOU AND A 5 STAR VOTE FOR YOUR ANSWER PLZ HELP
tensa zangetsu [6.8K]

Answer:

80, every 5 percent is 4 people. So every 25% is 20. 20 x 4 is 80.

Step-by-step explanation:

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What is the unit price of a quart of milk for $0.89?
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$3.65/gallon APEX

Step-by-step explanation:

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