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3 years ago

Use the coordinates to find the length of each side of the rectangle. Then find the perimeter. Q(0,0),R(4,0), S(4,4), T(0,4)

1 answer:

4 0

The length of each side of the square actually is 4 units. The area is 16 units squared and the perimeter is is ALSO 16.

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3 years ago

Help me to solve this.<br><br>x + 5 = 5

Ivenika [448]

The answer is x=0

Step by step explanation:
X+5=5
X+5-5=5-5
X=0.

7 0

3 years ago

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If there is a 75 x 50 hardwood top and bottom rail in this fence to connect the posts, how many 5.4 m long lengths of timber rai

kondaur [170]

Answer:

3,750

Step-by-step explanation:

because you need to do 7 * 75 * 50 and you will get the answer but it will be 3007 715 so that would be 7 3750h fences

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3 years ago

A can in the shape of a cylinder has a diameter of 6 centimeters and a height of 10 centimeters. Which measurement is closest to

VladimirAG [237]

Answer:

245.04

Step-by-step explanation:

A=2πrh+2πr2=2·π·3·10+2·π·32≈245.04423

u have to divide 6 and 2 to find the radius

hope this helps

7 0

3 years ago

Read 2 more answers

Select the point that is a solution to the system of inequalities

Zarrin [17]

<h2>Hello!</h2>

The answer is:

The point that is a solution to the system of inequalities is C(4,2).

To find the point that is a solution to the system of inequalities, we need to evaluate it into the given inequalities. If the point is a solution to the system of inequalities, both inequalities will be satisfied.

We are given the inequalities:

$y\leq 2x-2$

$y\leq x^{2} -3x$

So, substituting the given points into the given inequalities, we have:

- A. (2,1)

Substituting into the first inequality, we have:

$y\leq 2x-2\\\\1\leq 2*(2)-2\\\\1\leq 4-2\\\\1\leq 2$

Substituting into the second inequality, we have:

$y\leq x^{2} -3x$

$1\leq (2)^{2} -3(2)$

$1\leq 4 -6$

$1\leq -2$

Therefore, since 1 is not less or equal to -2, the point A(2,1) is not a solution to the system of inequalities.

- B. (-2,-1)

Substituting into the first inequality, we have:

$y\leq 2x-2\\\\-1\leq (-2)*(2)-2\\\\-1\leq -4-2\\\\-1\leq -6$

Substituting into the second inequality, we have:

$y\leq x^{2} -3x$

$-1\leq (-2)^{2} -3(-2)$

$-1\leq 4 +6$

$-1\leq 10$

Therefore, since -1 is not less or equal to -6, the point B(-2,-1) is not a solution to the system of inequalities.

- C. (4,2)

Substituting into the first inequality, we have:

$y\leq 2x-2\\\\2\leq (4)*(2)-2\\\\2\leq 8-2\\\\2\leq 6$

Substituting into the second inequality, we have:

$y\leq x^{2} -3x$

$2\leq (4)^{2} -3(4)$

$2\leq 16 -12$

$2\leq 4$

Therefore, since 2 is less than 6, and 2 is less than 4, the point B(4,2) is a solution to the system of inequalities.

- D(1,3)

Substituting into the first inequality, we have:

$y\leq 2x-2\\\\3\leq (1)*(2)-2\\\\3\leq 2-2\\\\3\leq 0$

Substituting into the second inequality, we have:

$y\leq x^{2} -3x$

$3\leq (1)^{2} -3(1)$

$3\leq 1 -3$

$3\leq -2$

Therefore, since 3 is not less than 0, and 3 is not less than -2, the point D(1.3) is not a solution to the system of inequalities.

Hence, the point that is a solution to the system of inequalities is C(4,2).

Have a nice day!

6 0

3 years ago