MATHEMATICS MIDDLE SCHOOL

The sides of a square are 3 cm long. One vertex of thesquare is at (2,0) on a square coordinate grid marked in
centimeter units. Which of the following points could
also be a vertex of the square?
F. (−4, 0)
G. ( 0, 1)
H. ( 1,−1)
J. ( 4, 1)
K. ( 5, 0)

Answers

Answer 1
Answer:

Answer:  The required point that could also be a vertex of the square is K(5, 0).

Step-by-step explanation:  Given that the sides of a square are 3 cm long and one vertex of the  square is at (2,0) on a square coordinate grid marked in  centimeter units.

We are to select the co-ordinates of the point that could also be a vertex of the square.

To be a vertex of the given square, the distance between the point and the vertex at (2, 0) must be 3 cm.

Now, we will be suing the distance formula to calculate the lengths of the segment from the point to the vertex (2, 0).

If the point is F(-4, 0), then the length of the line segment will be

\ell=√((-4-2)^2+(0-0)^2)=√(6^2+0^2)=√(6^2)=6~\textup{cm}\neq 3~\textup{cm}.

If the point is G(0, 1), then the length of the line segment will be

\ell=√((0-2)^2+(1-0)^2)=√(2^2+1^2)=√(4+1)=\sqrt5~\textup{cm}\neq 3~\textup{cm}.

If the point is H(1, -1), then the length of the line segment will be

\ell=√((1-2)^2+(-1-0)^2)=√(1^2+1^2)=√(1+1)=\sqrt2~\textup{cm}\neq 3~\textup{cm}.

If the point is J(4, 1), then the length of the line segment will be

\ell=√((4-2)^2+(1-0)^2)=√(2^2+1^2)=√(4+1)=\sqrt5~\textup{cm}\neq 3~\textup{cm}.

If the point is K(5, 0), then the length of the line segment will be

\ell=√((5-2)^2+(0-0)^2)=√(3^2+0^2)=√(3^2)=3~\textup{cm}.

Thus, the required point that could also be a vertex of the square is K(5, 0).
Answer 2
Answer: K (5.0) It's easy just use 2plus3and that's it.

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Answers

Answer: 1.4 inches

Step-by-step explanation:

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96 - 94.6 = 1.4
The answer to your question is C

-11 = -5 + 6n + 6 WHAT DOES "N" EQUAL?

Answers

-11= -5+6n+6
-11+5-6=6n
-12=6n
n= -2

(hope i helped :D!)

The graph of which function has a minimum located at (4, –3)?f(x) = x2 + 4x – 11
f(x) = –2x2 + 16x – 35
f(x) = x2 – 4x + 5
f(x) = 2x2 – 16x + 35

Answers

Answer:

Step-by-step explanation:

In order to solve the question, we have to derivate each function.

1) f(x) = x2 +4x -11

Then,

f'(x)= 2x +4

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 2*4 +4 = 12 ≠ 0 then this function doesn't not have a minimum at (4, -3)

2) f(x) = –2x2 + 16x – 35

Then,

f'(x)= -4x +16

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= -4*4 16 = 4 ≠ 0 then this function have a critical point at (4, -3)

then,

f''(4) =-4 <0 then we have a minimum at (4, -3)

3) f(x) = x2 – 4x + 5

Then,

f'(x)= 2x -4

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 2*4 -4 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)

4) f(x) =  2x2 – 16x + 35

Then,

f'(x)= 4x -16

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 4*4 -16 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)
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Answers

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When do you carry the high number in vertical multiplication

Answers

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——-

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If y=x and y=x+2 are the same, then how many solutions does it have

Answers

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Step-by-step explanation:

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There are no values of x or y that make this true.

_____

The two equations describe parallel lines, so they cannot be "the same." There are no points of intersection.
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