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slavikrds [6]
3 years ago
8

PLEASE HELP!!!! f(x)= {____for____ {____for____

Mathematics
1 answer:
hammer [34]3 years ago
8 0

9514 1404 393

Answer:

  f(x)=\left\{\begin{array}{ccc}3&\text{for}&-5\le x < 0\\1-x&\text{for}&0

Step-by-step explanation:

The left section is defined at x = -5, but not at x = 0, so the interval for that definition is -5 ≤ x < 0. The function value on that interval is constant: 3.

The right section is defined a x = 5, but not at x = 0, so the interval for that definition is 0 < x ≤ 5. That section is a straight line with a slope of -1 and a y-intercept of 1. The function value on that interval can be written -x+1.

Then the piecewise definition is ...

  f(x) = {3, for -5 ≤ x < 0; -x+1, for 0 < x ≤ 5}

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B i think let me solve

Step-by-step explanation:

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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

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

\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>

6 0
3 years ago
What is the solution to this equation?
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Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
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Lorico [155]

Answer:

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Step-by-step explanation:

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/7        /7

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6 0
3 years ago
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