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ale4655 [162]
3 years ago
15

Find an equation of the circle that has center (-3,3) and passes through (6,-2)

Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

( x + 3)^2 + ( y - 3)^2 = 106

Step-by-step explanation:

The equation of the circle with a center and a point

( x - a) ^2 + ( y - b) ^2 = r^2

( -3 , 3) - center - (a, b)

a = -3

b = 3

( 6 , -2) - point - ( x, y)

x = 6

y = -2

Step 1: substitute the center into the equation

( x -(-3)^2 + ( y - 3)^2 = r^2

( x + 3)^2 + ( y - 3)^2 = r^2

Step 2: sub the point into the equation

( 6 + 3)^2 + ( -2 - 3)^2 = r^2

9^2 + ( -5)^2 = r^2

81 + 25 = r^2

106 = r^2

Step 3: sub the radius into the equation

( x + 3)^2 + ( y - 3)^2 = 106

The equation of the circle is

( x + 3)^2 + ( y - 3)^2 = 106

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the cylinder has radius 4√3 and hight h. the total surface area of the cylinder is 56π√6. find the exact value of h giving the a
Arte-miy333 [17]

The area of the cylinder is a function of its height (h) and radius, (\mathbf{4\sqrt 3})

The exact value of h is: \mathbf{7\sqrt 2- 4\sqrt 3}

The given parameters are:

\mathbf{Area =56\pi\sqrt 6}

\mathbf{r=4\sqrt 3}

The surface area of a cylinder is calculated as:

\mathbf{Area = 2\pi rh + 2\pi r^2}

Substitute values for Area

\mathbf{56\pi\sqrt 6= 2\pi rh + 2\pi r^2}

Divide through by pi

\mathbf{56\sqrt 6= 2 rh + 2r^2}

Substitute value for r

\mathbf{56\sqrt 6= 2 (4\sqrt 3)h + 2(4\sqrt 3)^2}

\mathbf{56\sqrt 6= 8h\sqrt 3 + 2\times 48}

\mathbf{56\sqrt 6= 8h\sqrt 3 + 96}

Collect like terms

\mathbf{8h\sqrt 3 = 56\sqrt 6- 96}

Make h the subject

\mathbf{h = \frac{56\sqrt 6}{8\sqrt 3}- \frac{96}{8\sqrt 3}}

\mathbf{h = 7\sqrt 2- \frac{12}{\sqrt 3}}

Rationalize

\mathbf{h = 7\sqrt 2- \frac{12\sqrt 3}{3}}

\mathbf{h = 7\sqrt 2- 4\sqrt 3}

Hence, the exact value of h is: \mathbf{7\sqrt 2- 4\sqrt 3}

Read more about surface areas t:

brainly.com/question/25131428

5 0
2 years ago
Ln a+3 + ln a-3 = ln 16
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\bf log_{{  a}}(xy)\implies log_{{  a}}(x)+log_{{  a}}(y)\\\\
and\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
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ln[a^2-3^2]=ln(16)\impliedby \textit{removing ln() from both sides}
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3 years ago
Which statement is true of triangles P and Q? Triangle P has side lengths of 6, 8, 10 and angle measures of 53.1 degrees, 90 deg
Wewaii [24]

Answer:

Triangle P and Triangle Q are mathematically similar shapes (?).

Step-by-step explanation:

Hi, so the question asks which statement is true, given the following information, but you haven't written what statements we can choose from.

After reading the information, we can see that Triangle Q is the same shape as Triangle P but just larger.

I'm assuming that one of the statements given is about Triangle P and Triangle Q being mathematically similar shapes?

If you need to show your working out, here it is:

18 ÷ 6 = 3

24 ÷ 8 = 3

30 ÷ 10 = 3

All the angles are the same.

This means that the length scale factor is +3 from Triangle P to Triangle Q, the area scale factor is +9 (because 3 x 3 = 9) from Triangle P to Triangle Q, and that the two shapes are mathematically similar.

*DISCLAIMER* The majority of question askers on Brainly seem to be from the US, and I'm not, so the way I work things out / the mathematical terms I use might be different. Sorry!

Hope this helped anyway!

Bluey :)

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