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3 years ago

Draw two possible rectangles for a perimeter of 8 units

Mathematics

2 answers:

Kitty [74]3 years ago

7 0

All four side being 2 units (2+2+2+2=8)

or

2 sides being 1 and the other 2 side being 3 (1+1+3+3=8)

Darina [25.2K]3 years ago

7 0

When you draw it 1 rectangle each side is 1 centimeter or inch depends on what youre measuring it in.

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Emily quiz scores for this semester were 82,73,73,91 and 85 what is the

Elanso [62]

The median is 82
The mode is 73
The mean is 80.8

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3 years ago

In ΔEFG, the measure of ∠G=90°, FG = 15 feet, and GE = 87 feet. Find the measure of ∠F to the nearest degree

Ostrovityanka [42]

Answer:

Step-by-step explanation:

delta math

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3 years ago

Read 2 more answers

5. The distance between two given points (5,2) and (x,5) is 5 units. Find the possible values of x. *

kirza4 [7]

<h3>Answers: x = 1 and x = 9</h3>

============================================================

Explanation:

We'll use the distance formula here. Rather than compute the distance d based on two points given, we'll go in reverse to use the given distance d to find what the coordinate must be to satisfy the conditions.

We're given that d = 5

The first point is $(x_1,y_1) = (5,2)$ and the second point has coordinates of $(x_2,y_2) = (x,5)$ where x is some real number.

We'll plug all this into the distance formula and solve for x.

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\5 = \sqrt{(5-x)^2+(2-5)^2}\\\\5 = \sqrt{(5-x)^2+(-3)^2}\\\\5 = \sqrt{(5-x)^2+9}\\\\\sqrt{(5-x)^2+9} = 5\\\\(5-x)^2+9 = 5^2\\\\(5-x)^2+9 = 25\\\\(5-x)^2 = 25-9\\\\(5-x)^2 = 16\\\\5-x = \pm\sqrt{16}\\\\5-x = 4 \ \text{ or } \ 5-x = -4\\\\-x = 4-5 \ \text{ or } \ -x = -4-5\\\\-x = -1 \ \text{ or } \ -x = -9\\\\x = 1 \ \text{ or } \ x = 9\\\\$

This means that if we had these three points

A = (5, 2)
B = (1, 5)
C = (9, 5)

Then segments AB and AC are each 5 units long.

7 0

3 years ago

What is .the quotient -5 and y ??

stepladder [879]

Answer:

5 is = to 6 . 4 is equal to x so y is 64 divided bye 5 witch is 12

Step-by-step explanation:

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3 years ago

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$

$\text{The length of the longer diagonal}=4+9.54$

$\text{The length of the longer diagonal}=13.54$

Therefore, the length of longer diagonal is approximately 13.54 inches.

3 0

3 years ago