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Leviafan [203]
3 years ago
14

What are the x-intercepts of the quadratic function?

Mathematics
2 answers:
maw [93]3 years ago
8 0
To find the x intercept set y = 0

x^2 + x  - 2 =0 

(x+2)(x-1) = 0

x = -2 , x = 1


Anna71 [15]3 years ago
8 0
Your answer should be the last one, -2 and 1. Hope this helps!
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Can anyone do this 3 questions please....
Colt1911 [192]

Answer:

23 ) 60 x raised to 3  , y raised 3  , z raised to 3

25 ) 12 p

Step-by-step explanation:

23 ) volume of a rectangular box = l ×b×h

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        60 x raised to 3

25 )  area of a square = side × side

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5 0
3 years ago
14. Order the following from least to greatest. Angles ∠B, ∠C, ∠BDC. Sides DE, EF, and DF (Choose one for Angles and one for Sid
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3 0
3 years ago
Multiple by moving the decimal point 24.6x100,37x1,000 0.367x10,000,0.005x1000,000 .
lesya [120]

The answers are 2460, 37,000, 3670 and 5000

Step-by-step explanation:

  • Step 1: Find the answers for each by multiplying.

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37 × 1000 = 37,000

0.367 × 10000 = 3670 (moved three decimal points)

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8 0
4 years ago
Luke can paint 91 portraits in 7 weeks. how many portraits does he paint each week?
shutvik [7]

Answer:

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7 0
3 years ago
What is the equation of a parabola with (−2, 4) as its focus and y = 6 as its directrix? Enter the equation in the box.
SOVA2 [1]
Notice the picture below

the directrix is above the focus point, meaning the parabola is vertical and is opening downwards

now, "p" is the distance from the vertex to the focus point or the directrix, so that means, the vertex is between those two fellows, over the axis of symmetry, x = -2, since "p" is 1 unit, that puts the vertex at -2,5

since the parabola is opening downwards, that means the "p" value is negative, so is -1

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-----------------------------\\\\

\bf \begin{cases}
p=-1\\
h=-2\\
k=5
\end{cases}\implies (x-(-2))^2=4(-1)(y-5)
\\\\\\
(x+2)^2=-4(y-5)\implies 
-\cfrac{1}{4}(x+2)^2=y-5
\\\\\\
\boxed{-\cfrac{1}{4}(x+2)^2+5=y}

4 0
3 years ago
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