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Leno4ka [110]
3 years ago
7

What is a value or values that make an equation true

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
7 0
Any value that satisfies the equality.
E.g. 2x + 4x = 10 the value that makes this true is 2

or 

x^2 = 2 The values that would make it true are 4 and -4, since the squared of any number is always a positive.
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13-6x=(2x-5)^2+3 what answer
svet-max [94.6K]
<span><span>13−<span>6x</span></span>=<span><span><span>(<span><span>2x</span>−5</span>)</span>2</span>+3</span></span>Step 1: Simplify both sides of the equation.<span><span><span>−<span>6x</span></span>+13</span>=<span><span><span>4<span>x2</span></span>−<span>20x</span></span>+28</span></span>Step 2: Subtract 4x^2-20x+28 from both sides.<span><span><span><span>−<span>6x</span></span>+13</span>−<span>(<span><span><span>4<span>x2</span></span>−<span>20x</span></span>+28</span>)</span></span>=<span><span><span><span>4<span>x2</span></span>−<span>20x</span></span>+28</span>−<span>(<span><span><span>4<span>x2</span></span>−<span>20x</span></span>+28</span>)</span></span></span><span><span><span><span>−<span>4<span>x2</span></span></span>+<span>14x</span></span>−15</span>=0</span>Step 3: Use quadratic formula with a=-4, b=14, c=-15.<span>x=<span><span><span>−b</span>±<span>√<span><span>b2</span>−<span><span>4a</span>c</span></span></span></span><span>2a</span></span></span><span>x=<span><span><span>−<span>(14)</span></span>±<span>√<span><span><span>(14)</span>2</span>−<span><span>4<span>(<span>−4</span>)</span></span><span>(<span>−15</span>)</span></span></span></span></span><span>2<span>(<span>−4</span>)</span></span></span></span><span>x=<span><span><span>−14</span>±<span>√<span>−44</span></span></span><span>−<span>8</span></span></span></span>
5 0
3 years ago
Write an equation of the circle that has a diameter with endpoints (12, 3) and<br> (-18,3).
Mademuasel [1]

The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0

<h3>Coordinates of center of circle</h3>

Since the endpoints of the diameter are (x₁, y₁) = (12,3) and (x₂, y₂) = (-18,3), the coordinates of the center of the circle are the midpoints of the diameter. So, the midpoints are

  • x = (x₁ + x₂)/2 = (12 + (-18))/2 = (12 - 18)/2 = -6/2 = -3 and
  • y = (y₁ + y₂)/2 = (3 + 3)/2 = 6/2 = 3.

So, the coordinates of the center of the circle are (-3, 3)

<h3>The radius of the circle</h3>

The radius of the circle r = √[(x₁ - h)² + (y₁ - k)²] where

  • (x₁, y₁) = coordinates of end of diameter = (12, 3) and
  • (h, k) = coordinates of center of circle = (-3, 3)

So, substituting the values of the variables into r, we have

r = √[(x₁ - h)² + (y₁ - k)²]

r = √[(12 - (-3))² + (3 - 3)²]

r = √[(12 + 3)² + 0²]

r = √[15² + 0²]

r = √15²

r = 15

<h3>The equation of the circle</h3>

The equation of a circle with center (h,k) is given by

(x - h)² + (y - k)² = r² where r = radius of circle.

Substituting the values of the variables into the equation, we have

(x - h)² + (y - k)² = r²

(x - (-3))² + (y - 3)² = 15²

(x + 3)² + (y - 3)² = 15²

x² + 6x + 9 + y² - 6y + 9 = 225

x² + 6x + y² - 6y + 9 + 9 - 225 = 0

x² + y² + 6x - 6y - 207 = 0

The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0

Learn more about equation of a circle here:

brainly.com/question/18435467

5 0
2 years ago
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