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4 years ago

Which expression is equivalent to 3x+10−x+12 ? 26x 24x 2x + 22 4x + 22

Mathematics

1 answer:

Zina [86]4 years ago

4 0

Answer: 2x + 22

Solution: Simplify the expression.

3x + 10 - x + 12

3x - x = 2x

10 + 12 = 22

Therefore, the answer is 2x + 22.

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What is the rate of growth of f(x)= 3(1.04)x

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Answer:312 i thin

Step-by-step explanation:

3x1.04 then the =312 i think 312

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3 years ago

325 people were in a tour group. Six buses were completely filled and 7 people chose to walk. How many people did the bus hold?

anyanavicka [17]

First subtract 325-7= 318 Then you divide 318/6=53

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3 0

3 years ago

Read 2 more answers

Square RSTU is inscribed in circle P. Given the coordinates for the vertices of the square, find the equation of circle P.

svlad2 [7]

Answer:

$(x-2)^2+y^2=20$

Step-by-step explanation:

We want to find the equation of a circle P, and we know that the square RSTU is inscribed in the circle, this means that if we replace the points:

R(0,4) S(6,2) T(4,-4) and U(-2,-2) the equation must be verified.

Then, we have to replace the points in each option to see which one is the correct answer.

Option A: $(x-2)^2+y^2=68$

R(0,4):

$(x-2)^2+y^2=68\\(0-2)^2+4^2=68\\(-2)^2+16=68\\4+16=68\\20\neq 68$

We can see that R doesn't belong to the equation of the circle.

We could replace the rest of the points in the equation, but since we already know that R does not belong to the circle, we already know that it is not the right option.

Option B: $(x+2)^2+y^2=20$

R(0,4):

$(x+2)^2+y^2=20\\(0+2)^2+4^2=20\\2^2+4^2=20\\4+16=20\\20=20$

The point R(0,4) verifies the equation.

S(6,2):

$(x+2)^2+y^2=20\\(6+2)^2+2^2=20\\8^2+4^2=20\\64+16=20\\80\neq 20$

The point S doesn't belong to the equation of the circle, then we already know that this isn't the correct option.

Option C: $(x-2)^2+y^2=20$

R(0,4):

$(x-2)^2+y^2=20\\(0-2)^2+4^2=20\\(-2)^2+16=20\\4+16=20\\20=20$

The point R verifies the equation.

S(6,2):

$(x-2)^2+y^2=20\\(6-2)^2+2^2=20\\4^2+4=20\\16+4=20\\20=20$

The point S verifies the equation.

T(4,-4):

$(x-2)^2+y^2=20\\(4-2)^2+(-4)^2=20\\2^2+16=20\\4+16=20\\20=20$

The point T verifies the equation.

U(-2,-2):

$(x-2)^2+y^2=20\\(-2-2)^2+(-2)^2=20\\(-4)^2+4=20\\16+4=20\\20=20$

The point U verifies the equation.

We can see that the four points RSTU verifies the equation. This means that the square RSTU is inscribed in circle P whose equation is: $(x-2)^2+y^2=20$

We can see that the graph of the equation verifies the answer.