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Rom4ik [11]
3 years ago
15

Two students factored 2x^2 + 6x - 20. Leuko said that the factorization was (2x - 4)(x + 5). Ray gave the factorization as (x -

2)(2x + 10). Confirm that both of these factorizations are correct. Then explain why they are not complete.
Mathematics
2 answers:
shutvik [7]3 years ago
6 0

Answer:

See explanation

Step-by-step explanation:

If you FOIL both of them, you see that they are both correct, but they both need to factor out a 2 and then they will both have the same thing.

2(x - 2)(x + 5)

stich3 [128]3 years ago
5 0

Answer: To confirm the equations to be equal with the parent function we do as follows:

(2x – 4)(x + 5) = 2x^2 + 10x - 4x - 20 = 2x^2 + 6x -20

(x – 2)(2x + 10) = 2x^2 +10x - 4x -20 = 2x^2 +6x - 20

Step-by-step explanation:  Mark me as brainliest

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4 years ago
Roberto watched 5 more hours of tv than Jamie last week. Together, they watched a total of 23 hours of tv last week. How many ho
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Answer: 14 hours

Step-by-step explanation:

Let the number of hours of TV that Jamie watched be represented by x.

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3 0
3 years ago
A hyperbola centered at the origin has a vertex at (9, 0) and a focus at (−15, 0). Which are the equations of the asymptotes? y
chubhunter [2.5K]
Well, we know is centered at the origing, thus h,k are just 0,0.

using the provided vertex and focus point, gives us a distance for "a" of 9 and a distance for "c" of 15, check the picture below.

\bf \textit{hyperbolas, horizontal traverse axis }\\\\&#10;\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ( h, k)\\&#10;vertices\ ( h\pm a,  k)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{ a ^2+ b ^2}\\&#10;\textit{asymptotes}\quad  &#10;y= k\pm \cfrac{b}{a}(x- h)&#10;\end{cases}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;a=9\\&#10;c=15&#10;\end{cases}\implies c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b&#10;\\\\\\&#10;\sqrt{15^2-9^2}=b\implies \sqrt{144}=b\implies \boxed{12=b}\\\\&#10;-------------------------------\\\\&#10;asymptotes\qquad y=0\pm\cfrac{12}{9}(x-0)\implies y=\pm\cfrac{4}{3}x

3 0
4 years ago
Read 2 more answers
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