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3 years ago

Please help!!

Mathematics

2 answers:

Naya [18.7K]3 years ago

5 0

Answer:

Step-by-step explanation:

it is b

san4es73 [151]3 years ago

3 0

I worked it out it’s B

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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.

3 0

4 years ago

The surface area of this triangular pyramid, whose base and lateral faces are congruent equilateral triangles, is __ square inch

DedPeter [7]

Answer:

The surface area is equal to $390\ in^{2}$

Step-by-step explanation:

The surface area of the triangular pyramid is equal to the area of its four triangular faces

In this problem

$SA=4[\frac{1}{2}(b)(h)]$

we have

$b=15\ in$

$h=13\ in$

substitute the values

$SA=4[\frac{1}{2}(15)(13)]=390\ in^{2}$

4 0

3 years ago

Read 2 more answers

Someone please help me ASAP!

PilotLPTM [1.2K]

Answer:

(x+1, y-3)

Step-by-step explanation:

Because moving x by positive one means that you shift the shape to the right, and subtracting the y by 3 means moving it down by three. I solved it by choosing one point and figuring out how the same point ended up where it is after the translation (I looked at point a and saw how much it moved right and down). Hope this helps! :)

3 0

3 years ago

Factor completely 15a^2b-10ab^2

ozzi

Find the Greatest Common Factor (GCF)

GCF = 5ab

Factor out the GCF (Write the GCF first. Then, in parentheses, divide each term by the GCF.)

5ab(15a^b/5ab + -10ab^2/5ab)

Simplify each term in parentheses

5ab(3a - 2b)

6 0

3 years ago

This is all the points I have left is anyone willing to help me?

Alik [6]

Answer:

Step-by-step explanation:

7 0

3 years ago