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4 years ago

Help center of circle

Mathematics

1 answer:

Lelu [443]4 years ago

7 0

Answer:

Center = (7, - 5)

Radius = 3

Step-by-step explanation:

${x}^{2} + {y}^{2} + 14x - 10y + 65 = 0 \\ {x}^{2} + 14x + {y}^{2} - 10 + 65 = 0 \\ ( {x}^{2} + 14x + {7}^{2}) - {7}^{2} + ({y}^{2} - 10y + {5}^{2}) \\- {5}^{2} + 65 = 0 \\ (x + 7)^{2} + (y - 5)^{2} - 49 - 25 + 65 = 0 \\ (x + 7)^{2} + (y - 5)^{2} - 49 + 40= 0 \\ (x + 7)^{2} + (y - 5)^{2} - 9= 0 \\ (x + 7)^{2} + (y - 5)^{2} = 9 \\ (x + 7)^{2} + (y - 5)^{2} = {3}^{2} \\ equating \: it \: with \: \\ (x - h)^{2} + (y - k)^{2} = {r}^{2} \\ - h = 7 \implies \: h = - 7 \\ - k = - 5 \implies \: k = 5 \\ center = ( - h, \: - k) = (7, \: \: - 5) \\ \\ radius \: r = 3$

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makkiz [27]

Solution :

Given :

2y''' + 15y'' + 24y' + 11y= 0

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$(2D^3+15D^2+24D+11)y=0$

Auxiliary equation,

$2m^3+15m^2+24m +11 = 0$

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The roots are

$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$

$=\frac{-13\pm9}{4}$

$=-5.5, -1$

So, $m_1, m_2, m_3 = -1, -1, -5.5$

Then the general solution is :

$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$

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3 years ago

Let f(x) = 4x + 3 and g(x) = -2x + 5. Find.(f o g) (5)

daser333 [38]

f(x) = 4x + 3
g(x) = -2x + 5

(f o g) (5) means the function of g when g is a function of 5
(the function of the function concept)

so we need to get g(5) firstly


g(x) = -2x + 5
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After that, f(g(5)) = f(-5) = 4(-5) + 3 = -20 + 3 = -17

I hope that helps

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yaroslaw [1]

Answer:

Results are below.

Step-by-step explanation:

Giving the following information:

Monthly deposit= $100

Interest rate= 0.06/12= 0.005

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a)

To calculate the future value, we need to use the following formula:

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Answer:

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Step-by-step explanation:

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($66.50 - 4y) / 3

x = (66.50 - 4y) / 3

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