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galina1969 [7]
3 years ago
7

Using the equation y = 2x + 5 , what would the input need to be for an output of -45?

Mathematics
2 answers:
Semenov [28]3 years ago
3 0

Answer:

-25

Step-by-step explanation:

Essentially, y is your output and x is your input.

Here's your base equation: y = 2x + 5

Knowing the information above, let's plug it in:

-45 = 2x + 5

(Subtract 5 from both sides)

-50 = 2x

(Divide both sides by 2 to isolate x)

-25 = x

And there you go.

marissa [1.9K]3 years ago
3 0

Answer:

input is -25

Step-by-step explanation:

y = 2x + 5

The output is -45 which is y

-45 = 2x + 5

Subtract 5 from each side

-45-5 = 2x

-50 = 2x

Divide each side by 2

-50/2 = 2x/2

-25 = x

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3 years ago
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
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\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

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