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snow_tiger [21]
3 years ago
8

Can someone answer this circle equation question? PLEASE SHOW STEPS

Mathematics
2 answers:
Soloha48 [4]3 years ago
4 0

Answer:

A

Step-by-step explanation:

oh steps are here!

ratelena [41]3 years ago
4 0

Answer:

a) 5

Step-by-step explanation:

2x² - 6x + 2y² + 2y = 45

Complete the square and get it in the form (x-a)² + (y -b)² = r²

2x² - 6x +  2y² + 2y  = 45

Divide the equation by 2

x² - 3x + y² + y = 45/2

x² -  3x + ? + y² + y + ?? = 45/2

? -----> (x coefficient by 2)² = (3/2)² = 9/4

?? ----> (y coefficient by 2)² = (1/2)² = 1/4

Add (9/4) & (1/4) to both sides,

<u>x² -3x +(9/4)</u> + <u>y² + y + (1/4)</u> = (45/2)+ (9/4) + (1/4)

(x - 3/2)² + (y + 1/2)² = (90/4) + (9/4) + (1/4)

(x-\frac{3}{2})^{2}+(y+\frac{1}{2})^{2}=\frac{90+9+1}{4}\\\\\\(x-\frac{3}{2})^{2}+(y+\frac{1}{2})^{2}=\frac{100}{4}\\\\\\(x-\frac{3}{2})^{2}+(y+\frac{1}{2})^{2}=25\\\\(x-\frac{3}{2})^{2}+(y+\frac{1}{2})^{2}=5^{2}

Radius  = 5

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Find x A. 212√2 B. 212–√ C. 213√2 D. 7
victus00 [196]

The value of x in the triangle is (d) 7

<h3>How to solve for x?</h3>

The complete question is in the attached image.

From the attached image of the triangle, we can see that the triangle is a right triangle, and x can be solved using the following sine function

\sin(45) = \frac{x}{7\sqrt 2}

Evaluate sin(45)

\frac 1{\sqrt 2} = \frac{x}{7\sqrt 2}

Solve for x

x = \frac{7\sqrt 2}{\sqrt 2}

Divide

x = 7

Hence, the value of x in the triangle is (d) 7

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7 0
2 years ago
Use calculus to find the area of the triangle with the vertices (0, 5), (2, -2), and (5, 1).
guajiro [1.7K]

The area of the triangle with the vertices (0, 5), (2, -2), and (5, 1) by using the calculus is  21 square unit.

We need to find the equation among all possible pairs and then integrate the equations from one co-ordinate to another co-ordinate

Equation of line passing through (0,5) and (2,-2) is

y-5 = [(-2-5)/(2-0)](x-0)

=>y-5 = (-7) /2x

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Equation of line passing through (0,5) and (5,1) is

y-5 = [(1-5) / (5-0)](x-0)

=>y-5 = (-4/5)x

=>y = (-4/5)x+5-------(eq2)

Equation of line passing from (2,-2) and (5,1) is

y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)

=>y+2=(3/3)(x-2)

=>y=x-4--------(eq3)

Now, we use definite integration to find the area between the different equation of line.

So, area enclosed between the equations is given by the

area =\int\limits^5_2[(-4/5)x+5 - (-7/2)x + 5)dx  + \int\limits^5_1[(-4/5)x+5 -(x-4)]dx

=>area=\int\limits^5_2(7/2-4/5)x dx + \int\limits^5_1((-9/5)x+9)dx

Using properties of integration,\int\limits x\, dx=x^{2}/2

=>area=\int\limits^5_2(27/10)x dx + \int\limits^5_1(-9/5)x+9)dx

=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)

=>area=(27/20)×(25-4) + (-9/5)×24+9×4

=>area = (27×21)/20 + (-216)/5+ 36

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=>area= [(567-261×4)+(36×20)]/20

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=>area=423/20

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4 0
1 year ago
Patricia purchased x meters of fencing. She originally intended to use all of the fencing to enclose a square region, but later
sergejj [24]

Answer:

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Step-by-step explanation:

For a perimeter length of x, the side of a square will be x/4 and its area will be (x/4)^2.

If one side of the square is shortened by y/2 and the adjacent side is lengthened by y/2, then the difference in side lengths will be y. The area of the resulting rectangle will be ...

  (x/4 -y/2)(x/4 +y/2) = (x/4)^2 -(y/2)^2

That is, the difference in area between the square and the rectangle is ...

  (x/4)^2 - ((x/4)^2 -(y/2)^2) = (y/2)^2 = y^2/4

The positive difference between the area of the square region and the area of the rectangular region is y^2/4 square meters.

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Paraphin [41]
Gravity simply does not allow that
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Line l contains the points (9, 6) and (-9, 3). give the slope m of any line perpendicular to l.
aivan3 [116]
The slope of a line with those points is 1/6m. To find a perpendicular slope find the reciprocal of the slope of the first line. The slope is 6m. 

Have a great day, brainliest would be fantastic
6 0
4 years ago
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