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Savatey [412]
2 years ago
9

What is the first step when rewriting y = –4x2 + 2x – 7 in the form y = a(x – h)2 + k?

Mathematics
1 answer:
sdas [7]2 years ago
3 0

Answer:

–4 must be factored from –4x2 + 2x

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Can someone solve this for me and tell me the answer
Zolol [24]
Z (>/=) -17 
start by subracting 26 then multiplying by 2
5 0
3 years ago
Read 2 more answers
6x - 2y = 5
laiz [17]

Answer:

The given system has NO SOLUTION.

Step-by-step explanation:

Here, the given system of equation is:

6 x -  2 y = 5     .......... (1)

3 x  - y = 10          .... (2)

Multiply equation 2 with (-2), we get:

3 x  - y = 10     ( x -2)

⇒  - 6 x + 2 y = - 20

Now, ADD this to equation (1) , we get:

6 x - 2 y  - 6 x + 2 y  = 5 - 20

or, 0 = - 15

WHICH IS NOT POSSIBLE as 0 ≠ -15

Hence, the given system has NO SOLUTION.

7 0
3 years ago
A bicycle store costs ​$2250 per month to operate. The store pays an average of ​$40 per bike. The average selling price of each
lesantik [10]

Answer:

75

Step-by-step explanation:

Given That :

Operation cost = $2250 per month

Purchase price per bike = $40

Average selling price per bike = $70

How many bicycles must the store sell each month to break even ;

Break even point = point at which there is net profit / loss = 0

Let number of bicycles store must sell = x

Operation cost + (purchase price * Number of bicycles) = selling price * Number of bicycles

2250 + 40x = 70x

2250 = 70x - 40x

2250 = 30x

x = 2250 / 30

x = 75

To break even 75 bicycles must be sold

5 0
3 years ago
A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft
lakkis [162]

\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}

\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}

\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}

8 0
3 years ago
4w^3−8w+12<br> factor the polynomial<br><br> 5t^2+20t+50<br> factor the polynomial
DanielleElmas [232]
<span>4w^3−8w+12 = 4 (w^3 - 2w + 3)

</span><span>5t^2+20t+50 = 5 (t^2 + 4t + 10)</span>
7 0
3 years ago
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