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Temka [501]
3 years ago
15

Given the graph of the function f(x) below what happens to f(x) when x is a number between 0 and 1

Mathematics
2 answers:
GenaCL600 [577]3 years ago
5 0

Answer:

The answer would be D. Or F(x) is a negative number with a large absolute value.

Step-by-step explanation:

Apex

Andrews [41]3 years ago
4 0

Answer: correct answer is D.


Step-by-step explanation:


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Solve the equation for all values of x by completing the square.<br> 3x^2+177= -48x
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8

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4 0
3 years ago
if there is a 4 smiley faces in pattern 1 and 6 smiley faces in pattern 2 and 8 smiley faces in pattern 3 how many smiley faces
mixer [17]
We technically can't tell! Just because a pattern (increasing by two faces for every new pattern) is true at first doesn't mean it holds forever. However, if we make the assumption that the number of smiley faces continues to increase by 2, we have 4+87*2=177 as the answer, since we start with four and continue through a total of 87 more patterns, each of which adds two more faces.
4 0
3 years ago
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How are the expressions 0.4+1.5 and 1.5 +0.4 alike? different?
pav-90 [236]
They are equal to each other
5 0
3 years ago
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Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

As long as k\neq2 and k\neq6 (which we can't have because otherwise k^2-8k+12=0), we can cancel k^2-8k+12 in the denominators on both sides:

4=(k^2-6k)+(k-2)

4=k^2-5k-2

0=k^2-5k-6

We can factorize the right side:

0=(k-6)(k+1)

which tells us that k=6 and k=-1 are solutions.

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3 years ago
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