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Tanzania [10]
3 years ago
5

What is the first step when rewriting y = –4x2 2x – 7 in the form y = a(x – h)2 k? 2 must be factored from 2x – 7 –4 must be fac

tored from –4x2 2x x must be factored from –4x2 2x –4 must be factored from –4x2 – 7
Mathematics
1 answer:
liq [111]3 years ago
3 0

The first step in rewriting the function in a vertex form is (a) -4 must be factored from -4x² + 2x

<h3>How to determine the first step?</h3>

The equation is given as:

y = -4x² + 2x - 7

Factor out -4 from the first two terms of the equation

y = -4(x² + 0.5x) - 7

The above means that the first step in rewriting the function in a vertex form is (a) -4 must be factored from -4x² + 2x

Read more about vertex forms at:

brainly.com/question/18797214

#SPJ4

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You have botles of water how many can you buy with a 20$ and a bottle is 1.25​
igomit [66]

Answer:

16 bottles

Step-by-step explanation:

you can buy 16 bottles

7 0
4 years ago
I GIVE 5 STARS AND THANKS!! <br> Look at the screenshot--help??
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The store offers 10 type/color combinations. But, it can be 25 if you want to be technical .
4 0
3 years ago
Add or Subtract.<br> (−2^2 − 4 + 13) + (12^2 + 2 − 25)<br><br> (7^2 + 4 − 26) − (−7^2 − 3 + 15)
aleksklad [387]

Here is your answer

1. (−2^2 − 4 + 13) + (12^2 + 2 − 25)

= (4-4+13)+ (144+2-25)

[Note:- -2^2=(-2×-2)= +4]

= 13+ (146-25)

=13+121

= 134

2. (7^2 + 4 − 26) − (−7^2 − 3 + 15)

= (49+4-26) - (49-3+15)

= 49 +4 -26 -49 +3 -15

= 7-41

= -34

HOPE IT IS USEFUL

6 0
3 years ago
Which answer is the approximate standard deviation of the data set?
Ierofanga [76]

Answer:  A. 6.5

Step-by-step explanation:

Here, the given set = { 15, 19 3, 12, 17, 2, 2, 8}

Mean, \overline{ x} = \frac{15+19+3+12+17+2+2+8}{8} = \frac{78}{8} = 9.75

Number of elements, n = 8

\frac{\sum (|x-\overline{x}|)^2}{n}=42.4375

Thus, the standard deviation, \sigma =\frac{\sqrt{\sum(|x-\overline{x}|)^2} }{n}=\sqrt{\frac{339.5}{8}}=\sqrt{42.4375} = 6.51440711\approx 6.5

⇒ Option A is correct.

4 0
3 years ago
Read 2 more answers
How do you know where to put the constant when finding general solutions for differential equations?
Norma-Jean [14]
Let's suppose we want to solve y'=y with y(0)=2. Separating variables and integrating, we get

\displaystyle\int\dfrac{\mathrm dy}y=\int\mathrm dx\implies\ln|y|=x+C\implies y=e^{x+C}

Leaving the solution in this form, the initial condition gives

2=e^{0+C}=e^C\implies C=\ln2

This means the solution is y=e^{x+\ln2}.

Now if we were to write y=e^{x+C}=e^xe^C=Ce^x, then we would have found

2=Ce^0\implies C=2

so that the solution would have been y=2e^x.

But these two solutions are the same, since y=e^{x+\ln2}=e^xe^{\ln2}=2e^x. So we get the same solution regardless of where we place C, despite getting different values for C.
5 0
3 years ago
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