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3 years ago

if the xy-plane, the graph of y=x^2 and the circle with center (0,1) and radious 3 have how many points of intersection?

1 answer:

6 0

$\bf \textit{equation of a circle}\\\\
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
radius={{ r}}\\
----------\\
center\ (0,1)\to 
\begin{cases}
h=0\\
k=1
\end{cases}\\
r=3
\end{cases} 
\\\\\\
(x-0)^2+(y-1)^2=3^2\implies x^2+(y-1)^2=9
\\\\\\
(y-1)^2=9-x^2\implies y=\sqrt{9-x^2}+1\\\\
-----------------------------\\\\
\textit{where do the parabola and circle intersect? set them to equal}
\\\\\\

$

$\bf \begin{cases}
y=x^2\\\\
y=\sqrt{9-x^2}+1
\end{cases}\qquad y=y\implies x^2=\sqrt{9-x^2}+1$

solve for "x"

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<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

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2 years ago

Got a Home wrestling meet so wish me luck guys and gals

allsm [11]

Answer:

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Step-by-step explanation:

3 0

2 years ago

What is the volume of the cone with radius 6 m and height 12 m? Express the answer in terms of π.

Lina20 [59]

V = pi x R^2 x height
V= pi x 6^2 x 12
V= 432 pi (m^3)

6 0

2 years ago

Read 2 more answers

State the various transformations applied to the base function f(x)=|x| to obtain a graph of the function g(x) = |x| − 2.

Molodets [167]

Horizontal shift of 1 unit to the left and a vertical shift downward of 2 units.

<h3>Transformation of function</h3>

Transformation technique is a way of changing the position of an object on an xy-plane.

Given the parent function of a modulus function f(x) = |x|, the graph of the function g(x) = |x| - 2 shows a vertical translation of the parent function down by 2 units.

The resulting graph of the translated function is as shown below

Learn more on translation here: brainly.com/question/1046778

#SPJ1

6 0

1 year ago

Martin is cleaning the tires of his bicycle. He notices that his bicycle tire has a radius of 8 inches. How much bigger is the a

algol [13]

Strange question, as normally we would not calculate the "area of the tire." A tire has a cross-sectional area, true, but we don't know the outside radius of the tire when it's mounted on the wheel.

We could certainly calculate the area of a circle with radius 8 inches; it's

A = πr^2, or (here) A = π (8 in)^2 = 64π in^2.

The circumference of the wheel (of radius 8 in) is C = 2π*r, or 16π in.

The numerical difference between 64π and 16π is 48π; this makes no sense because we cannot compare area (in^2) to length (in).

If possible, discuss this situatio with your teacher.

8 0

2 years ago