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atroni [7]
2 years ago
11

Last oneeee helpppp pleaseeeee

Mathematics
2 answers:
tangare [24]2 years ago
7 0

13-4x = 13 -4(-8) = 13 +32 = 45

Rashid [163]2 years ago
4 0
<h2>☆ <em><u>Question</u></em> ☆</h2>

<h3>Simplify: 13 - 4x</h3><h3>when x = (-8)</h3><h2>_ _ _ _ _ _ _ _ _ _ _</h2><h2>☆ <em><u>Answer</u></em> ☆</h2>

=> 13 - 4x

=> 13 - 4(-8) ..(when x = (-8))..

=> 13 - (-32) ..( (-) , (-) = + )..

=> 13 + 32

=> 45

<h2>_ _ _ _ _ _ _ _ _ _ _</h2><h2>☆ <em><u>Final Answer</u></em> ☆</h2>

<h2>45</h2><h2>_ _ _ _ _ _ _ _ _ _ _</h2>

☆- PhysicsMaths - ☆

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a = ±10/9

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Find the center of a circle with the equation: x2 y2−32x−60y 1122=0 x 2 y 2 − 32 x − 60 y 1122 = 0
mixas84 [53]

The equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

The center of the circle exists at (16, 30).

<h3>What is the equation of a circle?</h3>

Let, the equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

We rewrite the equation and set them equal :

$(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2- 32x - 60y +1122=0

$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0

We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.

-2hx = -32x

h = -32/-2

⇒ h = 16.

-2ky = -60y

k = -60/-2

⇒ k = 30.

The center of the circle exists at (16, 30).

To learn more about center of the circle refer to:

brainly.com/question/10633821

#SPJ4

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2 years ago
I need help with the question !!
nikdorinn [45]

Answer:

Step-by-step explanation:

x = 136/2 = 68°

8 0
2 years ago
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