For two functions, m(x) and p(x), a statement is made that m(x) = p(x) at x = 7. What is definitely true about x = 7?

2 answers:

7 0

I answered this recently.

Choices were:
A)Both m(x) and p(x) cross the x-axis at 7.
B)Both m(x) and p(x) cross the y-axis at 7. 
C)Both m(x) and p(x) have the same output value at x = 7. 
D)Both m(x) and p(x) have a maximum or minimum value at x = 7.

The true statement was Choice C. Both m(x) and p(x) have the same output value at x = 7.

Alexus [3.1K]3 years ago

6 0

Answer:

Both m(x) and p(x) have the same output value at x = 7.

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Please help, I’ll mark your answer as brainliest!

statuscvo [17]

Answer:

$(x-230)^2+(y-220)^2=6100$

Step-by-step explanation:

The equation for a circle is $(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is the vertex of the circle and $r$ is the radius. Immediately, since the center of the circle is given, we know what $h$ and $k$ are. $h$ is 230 and $k$ is 220.

The only thing we need to find is the radius, which will just be the distance from the center (230,220) to a point on the circumference (170,170). The distance between them can be calculated using the distance formula, which is really just the Pythagorean Theorem rearranged. The formula states that $d=\sqrt{x^2+y^2}$ , where $x$ is the change in the x-coordinates of the two points and $y$ is the change in the y-coordinates of the two points. Plug-in $x$ for 60 and $y$ for 50 to get $d=\sqrt{50^2+60^2}$ . Solve for $d$ , arriving at $\sqrt{6100}$ . Therefore, the radius of the circle is $\sqrt{6100}$ .