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Ray Of Light [21]

1 year ago

14

I need help bad I dont understand this

1 answer:

Finger [1]1 year ago

4 0

Answer:

12.566 inches

Step-by-step explanation:

2πr/4

2π x8 divided by 4

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Helpppppppppppppppppppppp

Ad libitum [116K]

Answer:

8.04 units

Step-by-step explanation:

By Pythagoras Theorem:

${x}^{2} = \bigg( \frac{25.2}{2} \bigg)^{2} - {(9.7)}^{2} \\ \\ {x}^{2} = ( 12.6)^{2} - {(9.7)}^{2} \\ \\ {x}^{2} = 64.67 \\ \\ x = \sqrt{64.67} \\ \\ x = 8.04176598 \\ \\ x \approx \: 8.04$

4 0

2 years ago

Find the output, y, when the input, x , is 7.<br> y=

Anton [14]

The answer to your question would be 6!

4 0

3 years ago

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

andreyandreev [35.5K]

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.

3 0

3 years ago

(2x -2)^-3<br>can anyone help me with this?

zzz [600]

Answer: See below

Step-by-step explanation:

When you see that something is to the power of a negative number, it is another way of saying 1 over.

$(2x-2)^-^3=\frac{1}{(2x-2)^3} =\frac{1}{(2x-2)(4x^2-8x+4)} =\frac{1}{8x^3-8x^2+8x-8}$

4 0

3 years ago

Solve this problem.

Vesna [10]

Each interior angle = 540/5= 108

6 0

3 years ago