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nataly862011 [7]
2 years ago
5

Graph the following pair of quadratic functions and describe any similarities/differences observed in the graphs.

Mathematics
2 answers:
horsena [70]2 years ago
7 0

Answer:

The answer is a. both open upward; h is wider than f

Nikolay [14]2 years ago
3 0

Answer:

A

Step-by-step explanation:

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Find x. <br> I need help please
Misha Larkins [42]

Answer:

x = 4 \sqrt{3}

Step-by-step explanation:

By geometric mean theorem and Pythagoras theorem:

{x}^{2}  =  { (\sqrt{8 \times 4}  )}^{2} +  {4}^{2}  \\  \\  {x}^{2}  =  { (\sqrt{32}  )}^{2} +  {4}^{2}  \\  \\  {x}^{2}  =  32 + 16 \\  \\  {x}^{2}  = 48 \\  \\ x =  \sqrt{48}  \\  \\ x = 4 \sqrt{3}

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3 years ago
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cupoosta [38]

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1:6

Step-by-step explanation:

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3 0
2 years ago
Someone help me please and thank you!! ASAP :(
pantera1 [17]
<h3>Answer: Choice D) </h3>

Work Shown:

(f+g)(x) = f(x)+g(x)\\\\(f+g)(x) = \frac{x-23}{x^2+9x-36}+\frac{1}{x+12}\\\\(f+g)(x) = \frac{x-23}{(x+12)(x-3)}+\frac{1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+x-3}{x^2+9x-36}\\\\(f+g)(x) = \frac{2x-26}{x^2+9x-36}\\\\

We must require that x \ne -12 and x \ne 3 to avoid having 0 in the denominator. This is why choice D is the answer.

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3 years ago
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