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MaRussiya [10]
3 years ago
7

What are the solutions to the quadratic equation 4(x + 2)2 = 36 x = −11 and x = 7 x = −7 and x = 11 x = −5 and x = 1 x = −1 and

x = 5
Mathematics
2 answers:
Dmitriy789 [7]3 years ago
4 0

Answer:

c

Step-by-step explanation:

edge2020

Gnom [1K]3 years ago
4 0

Answer:

c

Step-by-step explanation:

source: trust me bro

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Answer:

Number line with numbers negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, 4 labeled. A dot is made at 2.6 and labeled as square root of 7

Step-by-step explanation:

* Lets explain how to solve this problem

- We want to know the position of √7 on the number line, then:

1- Square the √7

# (√7)² = 7

2- Search 7 is between which two square numbers

# 7 is between 4 and 9 which are square numbers

3- ⇒ 4 < 7 < 9 ⇒ take √ for all

# 2 < √7 < 3

∴ The value of √7 is between 2 and 3

* Lets look to the answer to find the correct answer

∵ The first answer is:

   A dot is made at 1.7 and labeled as square root of 7

∵ 1.7 is not between 2 and 3

∴ The first answer is not true

∵ The second answer is:

   A dot is made at 2.6 and labeled as square root of 7

∵ 2.6 is between 2 and 3

∴ The second answer is true

∴ Number line with numbers negative 4, negative 3, negative 2,

  negative 1, 0, 1, 2, 3, 4 labeled. A dot is made at 2.6 and labeled

  as square root of 7

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~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=3\\ k=-7 \end{cases}\implies y=a(x-3)^2-7\qquad \textit{we also know that} \begin{cases} x=1\\ y=-10 \end{cases} \\\\\\ -10=a(1-3)^2-7\implies -3=a(-2)^2\implies -3=4a\implies -\cfrac{3}{4}=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-\cfrac{3}{4}(x-3)^2-7~\hfill

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