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drek231 [11]
3 years ago
7

URGENT!! NEED THIS IN TWO HOURS. DUMB ANSWERS WILL BE REPORTED ALTHOUGH I UNDERSTAND IF YOU MAKE A SIMPLE MISTAKE IN YOUR CALCUL

ATIONS. I NEED THE FULL ANSWER AND EXPLANATION BELOW
Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).



You must include the appropriate sign (+ or -) in your answer. Do not use spaces in your answer.

8x^2 + 8y^2 _____x _______y _______= 0
ANSWERS MUST BE A ROUND NUMBER
Mathematics
1 answer:
Andre45 [30]3 years ago
6 0

Yo sup??

The given points are (1,1),(1,3) and (9,2)

equation of circle is

8x²+8y²+ax+by=0

let us plug in (1,1) in the equation we get

8+8+a+b=0

a+b=-16

let us now plug in (1,3) in the equation

8+8*3²+a+3b=0

a+3b=-80

let us. now solve the two equation in terms of a and b

we get

2b=--64

b=-32

and

a=-16+32

=16

therefore the equation is

8x²+8y²+16x-32y=0

Hope this helps

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How do you do the second part of the problem?
Sveta_85 [38]

Answer:

8

Step-by-step explanation:

According to the Alternating Series Estimation Theorem:

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7 0
4 years ago
How to differentiate ?
Bas_tet [7]

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

7 0
3 years ago
This is pretty easy :0
maksim [4K]

Answer:

1, 1.5, -1 1/5

Step-by-step explanation:

8 0
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trapecia [35]

Answer:

No is the answer:)

Step-by-step explanation:

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4 0
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