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4 years ago

The two points (-2,4) and (4,2) are the endpoints of the diameter of a circle. What is the equation of this circle in standard f

orm?

Mathematics

1 answer:

stepladder [879]4 years ago

3 0

$Center\ of\ the\ circle\ is\ the\ midpoint\ AB\ where\ A(-2;\ 4)\ \&\ B(4;\ 2)\\\\center:O\left(\frac{-2+4}{2};\ \frac{4+2}{2}\right)\to O(1;\ 3)\\\\Radius\ of\ the\ circle\ is\ distance\ O\ to\ A:\\\\r=|OA|=\sqrt{(-2-1)^2+(4-3)^2}=\sqrt{9+1}=\sqrt{10}\\\\Equation\ of\ a\ circle:\\\\(x-a)^2+(y-b)^2=r^2\ whete\ center\ is\ (a;\ b)\ and\ radius\ is\ r.\\\\Equation\ of\ this\ circle:\\\\(x-1)^2+(y-3)^2=(\sqrt{10})^2\\\\(x-1)^2+(y-3)^2=10$

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Polynomials <br><br> Multiply 5x(x2 - 4x + 2)

salantis [7]

Answer:

Assuming "x2" is x squared:

5x(x2-4x+2)

(5x(x2)+5x(-4x)+5x(2))

For these polynomial multiplication problems, just distribute the number outside the parentheses.

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3 years ago

Read 2 more answers

A lion needs to eat at least 20 pounds of food each

hram777 [196]

Answer:

3 more canisters.

Step-by-step explanation:

20 divided by 4 ( 4 pounds per container) is 5. When you divide 8 by 4, you get 2. 5-2=3. 3 more canisters! :) hope this helps!

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3 years ago

Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1

PolarNik [594]

Answer:

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

Step-by-step explanation:

We are given that function

$f(x)=(1+3x)^{-1/3}$

We have to find the first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

$f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....$

$f(0)=(1+3x)^{\frac{-1}{3}}=1$

$f'(x)=-\frac{1}{3}(1+3x)^{-\frac{4}{3}}(3)=-(1+3x)^{-\frac{4}{3}}$

$f'(0)=-1$

$f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}$

$f''(0)=4$

$f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}$

$f'''(0)=-28$

Substitute the values we get

$(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...$

$(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...$

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

5 0

3 years ago

The figure is made up of a cylinder and a hemisphere. To the nearest whole number, what is the approximate volume of this figure

Novosadov [1.4K]

Hello!

The figure is made up of a cylinder and a hemisphere. To the nearest whole number, what is the approximate volume of this figure?

Use 3.14 to approximate π .

Enter your answer in the box. in.³

Data: (Cylinder)

h (height) = 7 in

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: $\pi \approx 3.14$

V (volume) = ?

Solving: (Cylinder volume)

$V = \pi *r^2*h$

$V = 3.14 *2.5^2*7$

$V = 3.14*6.25*7$

$V = 137.375 \to \boxed{V_{cylinder} \approx 137.38\:in^3}$

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: $\pi \approx 3.14$

If: We know that the volume of a sphere is $V = 4* \pi * \dfrac{r^3}{3}$ , but we have a hemisphere, so the formula will be half the volume of the hemisphere $V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}$