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Vikki [24]
3 years ago
5

PLS HELP ILL GIVE BRAINLIEST

Mathematics
1 answer:
Blizzard [7]3 years ago
8 0

Hi there.

The answer is f(0) = 3.

Explanation:

Given 2 functions with given domains.

f(x) = 5x - 1

where x < -2

f(x) = x + 3

where x >= -2

Evaluate the value of f(x) when x = 0. Therefore, f(x)= 5x-1 cannot be used as the given x = 0 is greater than its domain.

Thus, our main function is f(x) = x+3

Evaluate the value by substituting x = 0 in the function.

f(0) = 0 + 3 \\ f(0) = 3

Thus, the answer is f(0) = 3

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Answer:

Kristina will!

Step-by-step explanation:

Both Krista and Kevin found the right answer to Stacey’s question. Stacey will invest $108 in company A if she invests $24 in company B.

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4 years ago
A seamstress has 1/4 of a foot of thread.Then she uses 1/5 of a foot of thread to sew a hole How much of the thread is left?
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1/10 of the thread is left because a connected to b is the thread and minus the 1/5 is 1/10 since 1/3 +total thread to begin wit.
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3 years ago
Read 2 more answers
Simultaneous Equations <br> 2x+4y=22 <br> 2x+2y=15
Aleks04 [339]

Answer:

x = 4

y = \frac{7}{2}

Step-by-step explanation:

Ummmm

I'm not sure if this is corrrect but this is what I got

So first you solve 2x+4y=22 to find x

and you get x = -2y+11

Then you have to subtitute -2y+11 for x in 2x+2y=15

which you then get y= \frac{7}{2}

After you have to subttitute y= \frac{7}{2}  for y in x = -2y+11

which you get as x=4

Finally you get the answer as:

x = 4

y = \frac{7}{2}

5 0
3 years ago
In Exercise,find the horizontal asymptote of the graph of the function.<br> f(x) = 16x/3+x^2
matrenka [14]

Answer:

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Step-by-step explanation:

I attached the graph of the function. Graphically it can be seen that the horizontal asymptote of the graph of the function is y=0.

When denominator's degree (2) is higher than nominator's degree (1) then the horizontal asymptote is y=0

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Please help with these questions :(
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In the Pythagorean theorem, a squared minim b squared equals c squared. If you apply that formule to question 2.

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