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Vlada [557]
3 years ago
12

-5m=4-6m I can't figure out the problem. I need help.

Mathematics
2 answers:
boyakko [2]3 years ago
8 0
-5m=4-6m: You would add -6m on both sides:
-5m=4-6m
+6m    +6m And then you would get 1m=4 then divide both sides by 1 and you would get m=4
r-ruslan [8.4K]3 years ago
5 0
-5m = 4 - 6m....add 6m to both sides
6m - 5m = 4
m = 4 <==
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What is the range of the function
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Answer: Choice B

Range = {-3, 1, 5}

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

Explanation:

The domain is the set of all possible input x values. The range is the set of all possible y outputs.

Plug in each x value from the domain, one at a time, to get its corresponding range y value.

--------------------

Start with x = -3

f(x) = 2x+3

f(-3) = 2(-3)+3

f(-3) = -6+3

f(-3) = -3

So -3 is in the range.

--------------------

Move onto x = -1

f(x) = 2x+3

f(-1) = 2(-1)+3

f(-1) = -2+3

f(-1) = 1

1 is also in the range

--------------------

Finally plug in x = 1

f(x) = 2x+3

f(1) = 2(1)+3

f(1) = 2+3

f(1) = 5

The value 5 is the final value in the range.

--------------------

All of those values form the set {-3, 1, 5} which is the complete range.

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2 years ago
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Step-by-step explanation:

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2 years ago
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In the equation (x^2+y)^5, what is the coefficient of the term x^4y^3? what is the coefficient of the same term in the expansion
lys-0071 [83]

\displaystyle&#10;(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

----------------------------------------------------

\displaystyle&#10;(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

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