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Evgesh-ka [11]
2 years ago
7

Which of the following equations will produce the graph shown below?

Mathematics
1 answer:
FromTheMoon [43]2 years ago
3 0

The equation that will produce the given graph is; 6x² + 6y² = 144

<h3>What is the equation of the Circle?</h3>

The standard equation of a circle can be expressed as:

(x + a)² + (y - b)² = r²

where;

(a, b) is the center of the circle

r is the radius of the circle.

From the diagram, we can see that the radius of the circle is between 4 and 5.  Thus, the correct equation must have a radius between these values.

From the option, we can see that the only equation that will have a radius between 4 and 5 is;

6x² + 6y² = 144

Divide through by 6 to get;

x² + y² = 24

x² + y² = √24

Thus r = √24 which is greater than 5

Read more about Equation of Circle at; brainly.com/question/1559324

#SPJ1

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Use the zero product property to find the solutions to the equation (x+2)(x+3)=12
Rainbow [258]

Answer:

<em>x=-6, x=1</em>

Step-by-step explanation:

<u>The zero product property</u>

It states that if a.b=0, then it must be satisfied that a=0 or b=0. It's commonly used to solve equations where one of the sides is 0.

The given equation is:

(x+2)(x+3)=12

Since neither side is 0, we operate the expression:

x^2+3x+2x+6=12

Moving 12 to the left side and simplifying:

x^2+3x+2x+6-12=0

x^2+5x-6=0

Factoring:

(x+6)(x-1)=0

Now, since there is zero on the right side, we apply the property to solve:

x+6=0, or x-1=0

The two solutions come directly:

x=-6, x=1

4 0
3 years ago
How to solve for x:
OLEGan [10]

Answer:

x = 2

Step-by-step explanation:

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2 years ago
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CaHeK987 [17]

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12 by 6 by 6

Step-by-step explanation:

3 0
3 years ago
A number has a 5 in the tens place. The number of ones is 3 less than the number of tens. The number of hundreds is 1 less than
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Well 3 less than 5 is 2 so we have 52 and 1 less than 2 is 1 so the answer is 152
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Please help <br><br> If f(x)=x^3+x-7 and g(x)=-5x, find (fog)(x) and (gof)(x)<br><br> (fog)(x)=?
Aleks04 [339]
I hope this helps you

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