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SIZIF [17.4K]
3 years ago
12

The function f shown in the graph is an even function.

Mathematics
1 answer:
Firlakuza [10]3 years ago
4 0

the top half is the bottom is not

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You and 4 of your friends want to make some soup for lunch each person would like to eat 1 1/3 cup soup how many cups of soup wi
Pavlova-9 [17]
Multiply the amount of people by the amount of cups they want


4*1 1/3=5 1/3
You would have to make 5 and 1/3 cups of soup

Hope this helps!
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3 years ago
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The center of a hyperbola is (−4,3) , and one vertex is (−4,7) . The slope of one of the asymptotes is 2.
Monica [59]

Answer:

The answer to your question is below

Step-by-step explanation:

C (-4, 3)

V (-4, 7)

asymptotes = 2 = \frac{b}{a}

- This is a vertical hyperbola, the equation is

       \frac{(y - k)^{2} }{a^{2} } + \frac{(x - h)^{2} }{b^{2} } = 1

slope = 2

a is the distance from the center to the vertex = 4

b = 2(4) = 8

       \frac{(y - 3)^{2} }{4^{2} } + \frac{(x + 4)^{2} }{8^{2} } = 1

       \frac{(y - 3)^{2} }{16} + \frac{(x + 4)^{2} }{64} = 1

7 0
3 years ago
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Solve each system by substitution.<br> -<br> y = 6x – 11<br> -2x – 3y=-7
densk [106]

Answer:

Point form (2.1)

Equation form: x=2,y=1

5 0
3 years ago
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What is the value of A when we rewrite... (PLZ HELP QUICK)
Marina86 [1]

Answer:

<h2>\frac{133}{8}</h2>

Step-by-step explanation:

Given,

{( \frac{5}{2} )}^{x}  +  {( \frac{5}{2} )}^{x + 3}

=  {( \frac{5}{2}) }^{x}  +  {( \frac{5}{2}) }^{x}  \times  {( \frac{5}{2} )}^{3}

= ( \frac{5}{2} ) ^{x} (1 +  {( \frac{5}{2} )}^{3}

=  {( \frac{5}{2} )}^{x} (1 +  \frac{125}{8} )

=  {( \frac{5}{2} )}^{x} ( \frac{1 \times 8 + 125}{8} )

=  {( \frac{5}{2}) }^{x} ( \frac{8 + 125}{8} )

{( \frac{5}{2} )}^{x} ( \frac{133}{8} )

Comparing with A • {( \frac{5}{2}) }^{x}

A = \frac{133}{8}

Hope this helps...

Good luck on your assignment...

7 0
3 years ago
Solve the proportion using equivalent ratios. Explain the steps you used to solve the proportion, and include the answer in your
NikAS [45]
10x=60
x=6
10/3=20/6
^^^^^^^^^^^^
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3 years ago
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