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3 years ago

Show that the equation represents a circle by rewriting it in standard form, and find the center and radius of the circle.

1 answer:

4 0

5x^2 + 5y^2 + 10x − y = 0 it is the same of 
x^2 + y^2 + 2x −1/5y = 0
and then 
(x + 1)^2 - 1+ (y−1/10)^2 -1/100=0
 (x + 1)^2 + (y −1/10)^2 =(101/100)

the center is C(-1, 1/10), and the radius is R= sqrt(101)/10

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X square= 121 can anyone solve it

Alik [6]

Answer:

Picture is the answer.

Step-by-step explanation:

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2 years ago

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BaLLatris [955]

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I believe the answer is y-axis

Step-by-step explanation:

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3 years ago

Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago

Please help. I'm stuck.

Bezzdna [24]

The root $\sqrt{10}$ can be converted into the power $10^{ \frac{1}{2} }$ . Therefore we can rewrite the problem as $(10^{ \frac{1}{2} } )^ {\frac{3}{4} x}$ and then follow the exponent rules about a power to a power, multiplying 1/2 and 3/4 together.

Thus the problem becomes $10^{\frac{3}{8}x}$ , which then can be turned into $\sqrt[8]{10} ^{3x}$ , making the last option our answer.

4 0

3 years ago

Please Help. 25 points.

vodomira [7]

Answer:

$\boxed{x = 7, y = 9, z = 68}$

Step-by-step explanation:

We must develop three equations in three unknowns.

I will use these three:

$\begin{array}{lrcll}(1) & 8x + 13y +7 & = & 180 & \\(2)& 9x - 7 + 13y +7 & = & 180 & \\(3)& 8x + 5y - 11 + z & = & 180 &\text{We can rearrange these to get:}\\(4)& 8x + 13y & = & 173 &\\(5) & 9x + 13y & = & 180 & \\(6)& 8x + 5y + z & = & 169 & \\(7)& x & = & \mathbf{7} & \text{Subtracted (4) from (5)} \\\end{array}$

$\begin{array}{lrcll}& 8(7) + 13y & = & 173 & \text{Substituted (7) into (4)} \\& 56 + 13y & = & 173 & \text{Simplified} \\& 13y & = & 117 & \text{Subtracted 56 from each side} \\(8)& y & =& \mathbf{9}&\text{Divided each side by 13}\\& 8(7) + 5(9) + z & = & 169 & \text{Substituted (8) and (7) into (6)} \\& 56 + 45 + z& = & 169 & \text{Simplified} \\& 101 + z& = & 169 & \text{Simplified} \\&z& = & \mathbf{68} & \text{Subtracted 101 from each side}\\\end{array}$

$\boxed{\mathbf{ x = 7, y = 9, z = 68}}$

4 0

3 years ago