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3 years ago

The circle below is centered at the point (4,1) and has a radius of length 2. What is it’s equation?

Mathematics

2 answers:

boyakko [2]3 years ago

8 0

Answer:

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k ) = (4, 1) and r = 2, hence

(x - 4)² + (y - 1)² = 4 → B

vekshin13 years ago

5 0

Answer:

Option B:

$(x-4)^2+(y-1)^2 = 4$

Step-by-step explanation:

Given

Centre = (h,k) = (4,1)

Radius = r = 2

The standard equation of circle is:

$(x-h)^2+(y-k)^2=r^2$

Putting the values of h,k and r

$(x-4)^2+(y-1)^2 = (2)^2\\(x-4)^2+(y-1)^2 = 4$

Hence, the correct option is option B

$(x-4)^2+(y-1)^2 = 4$

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Whats a frequency polygon

NeTakaya

Frequency polygons are a graphical device for understanding the shapes of distributions.

8 0

3 years ago

If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b

MAXImum [283]

We have

$\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}$

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0$

The rational term vanishes as x gets arbitrarily large, so we can ignore that term, leaving us with

$\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0$

and this happens if a = 1 and b = -1.

To confirm, we have

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0$

as required.

3 0

2 years ago

Which equation can you use to solve for x?

Nostrana [21]

X+71=108
because they're alternate angles and alternate angles have the same measures

7 0

3 years ago

In the inequality, what are all the possible values of x?

Marat540 [252]

Answer:

B) $\displaystyle x ≤ 0$

Step-by-step explanation:

2(x + 12) − 16 ≤ 8

$\displaystyle \frac{2(x + 12)}{2} ≤ \frac{24}{2} \\ \\ x + 12 ≤ 12 \\ \\ x ≤ 0$

I am joyous to assist you anytime.

7 0

3 years ago

What are the exact solutions of x2 + 5x + 1 = 0? (5 points)

Rus_ich [418]

Answer:

$x=\frac{-5\pm\sqrt{21} }{2}$

Step-by-step explanation:

We use the quadratic formula here which says for a quadratic equation $ax^2+bx+c=0$ .

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

Now in our case

$a=1\\b=5\\c=1$

so we have:

$x=\frac{-5\pm\sqrt{5^2-4(1)(1)} }{2(1)} =\frac{-5\pm\sqrt{21} }{2}$

$\boxed{x=\frac{-5\pm\sqrt{21} }{2} }$

Which are our solutions.

7 0

3 years ago