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BabaBlast [244]
3 years ago
6

Can someone answer this please!!

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

<u>The perimeter of the square is option E. 16√5 mm.</u>

<u>The area of the square is option A. 80 mm².</u>

Step-by-step explanation:

1. Let's calculate the perimeter of this square:

The perimeter of a square is the sum of its side 4 times, thus:

Perimeter = 4√5 + 4√5 + 4√5 + 4√5

<u>Perimeter = 16 √5</u>

<u>Perimeter = 4 * 4√5 = 16√5</u>

<u>The correct answer is option E. 16√5 mm</u>

2. Let's calculate the area of this square:

The area of a square is the length of its side squared, thus:

Area = (4√5)²

Area = 4√5 * 4√5

Area = (4 * 4 * √5 * √5)

Area = 16 * 5 = 80

<u>The correct answer is A. 80 mm²</u>

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The friend is wrong - According to the Pythagorean Theorm, if c is the hypotenuse and a and b are opposite sides then  c^{2} = a^{2}  +  b^{2} 
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raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci
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Answer:

Equation:

{x}^{2}   +  {y}^{2} +  2x  - 2y   -  35= 0

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )

Plug in the points to get:

( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )

( \frac{ -2}{2}, \frac{ 2}{2} )

(  - 1, 1)

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

r =  \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }

r =  \sqrt{ {(5 -  - 1)}^{2} + {( - 2- - 1)}^{2} }

r =  \sqrt{ {(6)}^{2} + {( - 1)}^{2} }

r =  \sqrt{ 36+ 1 }  =  \sqrt{37}

The equation of the circle is given by:

(x-h)^2 + (y-k)^2 =  {r}^{2}

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

(x- - 1)^2 + (y-1)^2 =  {( \sqrt{37}) }^{2}

(x + 1)^2 + (y - 1)^2 = 37

We expand to get:

{x}^{2}  + 2x  + 1 +  {y}^{2}  - 2y + 1 = 37

{x}^{2}   +  {y}^{2} +  2x  - 2y +2 - 37= 0

{x}^{2}   +  {y}^{2} +  2x  - 2y   -  35= 0

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

{x}^{2}   +  {0}^{2} +  2x  - 2(0)  -   35= 0

{x}^{2}   +2x   -   35= 0

(x - 5)(x + 7) = 0

x = 5 \: or \: x =  - 7

The point (5,0) and (-7,0) lies on the circle.

When x=0

{0}^{2}   +  {y}^{2} +  2(0)  - 2y   -  35= 0

{y}^{2} - 2y   -  35= 0

(y - 7)(y + 5) = 0

y = 7 \: or \: y =  - 5

The point (0,-5) and (0,7) lie on this circle.

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