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denis-greek [22]
2 years ago
8

The cost of silver is a function of the weight, what is the output when the input is 15

Mathematics
2 answers:
Over [174]2 years ago
4 0
The answer is 300 .
Vladimir79 [104]2 years ago
3 0

Answer:

rupee 300 will be the correct answer

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ulia spends $3.25 on gas for her lawn mower. She earns $14.00 mowing her neighbor's yard. What is Julia's profit?
ipn [44]
$14.00
- _3.25_
   $10.75
3 0
3 years ago
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I hate this, I’m not sure how to do these
Triss [41]

Answer:its B....sorry i didnt read the qn properly earlier

7 0
3 years ago
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g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and
allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [ab]_K is equal to [ba]_K, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [ab]_H = [ba]_H . Since a and b were generic elements of H, then H/G is abelian.

4 0
3 years ago
FIRST CORRECT ANSWER WILL GET BRAINLIEST
dezoksy [38]

Answer:

b

Step-by-step explanation:

vertical line through x at one

vertical lines have undefined slopes

can take two points to verify  (1,1)  (1,2)    m=<u> 2-1</u>

                                                                         1-1  

                                                                       <u>   1 </u>

                                                                          0         undefined zero in denominator

8 0
3 years ago
Evaluate $\dfrac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}}$.
Pavlova-9 [17]
<h3>Answer:  4</h3>

========================================================

Work Shown:

\frac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}}\\\\\frac{2\sqrt{36*2}}{\sqrt{4*2}+\sqrt{2}}\\\\\frac{2\sqrt{36}*\sqrt{2}}{\sqrt{4}*\sqrt{2}+\sqrt{2}}\\\\\frac{2*6*\sqrt{2}}{2*\sqrt{2}+\sqrt{2}}\\\\\frac{12\sqrt{2}}{2\sqrt{2}+\sqrt{2}}\\\\\frac{12\sqrt{2}}{3\sqrt{2}}\\\\\frac{12}{3}\\\\4

Note in step 2, I factored each number in the square root to pull out the largest perfect square factor. From there, I used the rule that \sqrt{A*B} = \sqrt{A}*\sqrt{B} to break up the roots.

8 0
2 years ago
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