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Olegator [25]
3 years ago
7

The circle below is centered at the point (3,1) and has a radius length of 2.What is its equation

Mathematics
1 answer:
hammer [34]3 years ago
7 0

Answer:

C. (x-3)^2+(y-1)^2=2^2

Step-by-step explanation:

The standard equation of a circle, given the radius r units and center (h,k) is given by:

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

From the question, we have the center of the circle to be (3,1) and the radius is 2 units.

We substitute the center and radius of the circle into the equation to get:

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

The correct answer is C

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Step-by-step explanation:

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What is the value of x, when 10(x + 2) = 5(x + 8)?
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7 0
3 years ago
If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0
alisha [4.7K]
x^2-1=x^2-1^2=(x-1)(x+1)

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark
8 0
3 years ago
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