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mina [271]
3 years ago
11

Given f(x)=4x+6 and g(x)=9x+9 then what is f(g(−6))?

Mathematics
2 answers:
irakobra [83]3 years ago
5 0
To solve this question, you can break it into 2 parts. First evaluate the function g(X)=9x+9 for g(-6). Which is g(-6) = 9(-6)+9 = -45. Then evaluate f(-45). F(-45)= 4(-45)+6= -180+6= -174. The final answer for f(g(-6))= -174.
Margarita [4]3 years ago
3 0
F(x)=4x+6
g(x)=9x+9f(g(−6))

f(x)·g(x) = (4x+6)(9x+9)
f(x)·g(x) = 36x^2 +36x+54x+54
f(x)·g(x) = 3<span>6x^2 +90x+54
</span>f(x)·g(x) = 3<span>6(-6)^2 +90(-6)+54
</span>f(x)·g(x) = 3<span>6(36)+90(-6)+54
</span>f(x)·g(x) = 1296<span>+(-540)+54
</span>f(x)·g(x) = 1296-540+54
f(x)·g(x) = 810
<span>f(g(−6)) = 810</span>
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Answer:

Answer C is correct.

Step-by-step explanation:

f(x) clearly has a maximum:  y = +10 at x = 0.

Analyzing g(x) = -(x + 1)^2 - 10, we see that the vertex is at (-1, -10), and that the graph opens down.  Thus, -10 is the maximum value; it occurs at x = -1.

Answer A is false.  Both functions have max values.

Answer B is false.  One max is y = 10 and the other is y = -10.

Answer C is correct.  The max value of f(x), which is 10, is greater than the max value of g(x), which is -10.

Answer D is false.  See Answer B, above.

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3 years ago
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Answer:

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2 years ago
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Y=mx+b
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y=2/3x-6
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9514 1404 393

Answer:

  300

Step-by-step explanation:

There are 25 ways to select the first student. After that student is removed from the selection pool for the second student, there are 24 ways to select the second student. This gives 25·24 = 600 ways to select 2 students <em>in a particular order</em>.

Since we don't care about the order, we can divide this number by the number of ways two students can be ordered: AB or BA, 2 ways.

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There are 300 ways to pick a combination of two students from 25.

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<em>Additional comments</em>

This sort of selection (2 out of 25) has a formula for it, and an abbreviation for the formula.

 "n choose k" can be written nCk or C(n, k)

The function is a ratio of factorials:

  nCk = n!/(k!(n-k)!)

If you can typeset this, it is written ...

  \displaystyle\binom{n}{k}=\frac{n!}{k!\cdot(n-k)!}

This is different from the formula for the number of <em>permutations</em> of n things taken k at a time. That would be written nPk or P(n, k) = n!/(n-k)!.

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the answer that 9-(2,4)

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