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morpeh [17]
3 years ago
12

Does the graph represent a function?

Mathematics
1 answer:
Schach [20]3 years ago
3 0

Answer: NO

Step-by-step explanation:

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The graph of F(x), shown below, resembles the graph of G(x)= x², but it has
NNADVOKAT [17]

The equation of the function is  F(x) = -3(x-4)² +4 , the correct answer is Option D

The missing graph is attached with the answer

<h3>What is a function ?</h3>

A function is a law that relate the independent variable and the dependent variable.

It is given that

The graph of F(x), shown below, resembles the graph of G(x)= x², but it has been changed

From the graph, it is seen that,

G(x) when shifted by -4 on the negative x axis.

and graph shifted by +4 on the positive y axis.

F(x) is scaled by a factor of -⅓ of G(x)

it is reflected across the x-axis.

So the equation of the function is

F(x) = -3(x-4)² +4

Therefore ,  the correct answer is Option D.

To know more about Function

brainly.com/question/12431044

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8 0
2 years ago
Sandy makes 15.5 gallons of lemonade. She must refill the container after filling 100 drink cups. How much lemonade was in each
faltersainse [42]
100 divided by 15.5 = 0.155
5 0
4 years ago
(b) A $150 bonus was awarded to the three salespeople, which they had to share. It was awarded so that each salesperson received
fiasKO [112]
Each person made $50.
150 divided by 3 = 50.
6 0
3 years ago
Find the coordinates of the point in the first quadrant at which the tangent line to the curve (x)^3-xy+(y)^3 =0 is parallel to
Oksanka [162]
<span>Differentiate implicitly:
</span>3x^2-y-xy'+3y^2y'=0
<span>
Solve for y
</span>y'(3y^2-x)=y-3x^2&#10;\\&#10;\\y'={y-3x^2\over3y^2-x}

<span>When the tangent is parallel to the x-axis we have y'=0, so we must solve
</span>y'={y-3x^2\over3y^2-x}=0\implies y=3x^2

<span>To find the actual value of x we plug this expression for y into the original equation
</span>x^3-3x^3+27x^6=0&#10;\\&#10;\\x^3(27x^3-2)=0\implies x=\{0,{\sqrt[3]2\over3}\}

<span>Plugging this into the formula for y above gives the points
</span>(0,0)\text{ and }({\sqrt[3]2\over3},{\sqrt[3]4\over3})

<span>which is where our tangent will be parallel to the x-axis.</span>
<span>

</span>
8 0
3 years ago
I have no idea how to do this, someone help me
julsineya [31]
Try this:
1) for square a=5*2=10.
for circle r=5.
2) for square S'=10²=100; for circle S°=πr²≈78.54.
3) for half of circle S₁=0.5*78.54=39.27
4) for shaded region S=S'-S₁=100-39.27≈60.73
Answer: G (60.73)
6 0
3 years ago
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