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Reika [66]
3 years ago
14

What is the center and radius of the circle with the given equation (x-1)^2+(y+1)^2=4?

Mathematics
2 answers:
melomori [17]3 years ago
8 0
Hi there! The centre of this circle is (1, -1) and the radius is 2.

The standard form of a circle's formula is
(x - a) {}^{2} + ( {y - b)}^{2} = r {}^{2}

With a and b representing the coordinates of the centre and r representing the radius of the circle. Since in this situation r^2 equals 4, the radius (r) has to be 2.
Mrrafil [7]3 years ago
4 0
<span>The center is x = 1, y = -1.
The radius is 2. </span>
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1/64 is between what two whole numbers
Ronch [10]

Answer:

Step-by-step explanation:

0 and 1

8 0
3 years ago
Laura has 24 stamps from each of 6 different countries. She can fit 4 stamps on each display sheets of an album. How many displa
Ket [755]
B. 36
To find the number of sheets that Laura is going to fill with her stamps, we first have to find the total number of stamps.
If she has 24 stamps EACH from 6 countries, to get the total, we will multiply
24 x 6 = 144 total number of stamps.

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144 ÷ 4 = 36
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3 0
3 years ago
Find f(x) and g(x) so that the function can be described as y = f(g(x)).<br><br> y = 4/x^2+9
dlinn [17]

I'm assuming all of (x^2+9) is in the denominator. If that assumption is correct, then,

One possible answer is f(x) = \frac{4}{x},  \ \ g(x) = x^2+9

Another possible answer is f(x) = \frac{4}{x+9}, \ \ g(x) = x^2

There are many ways to do this. The idea is that when we have f( g(x) ), we basically replace every x in f(x) with g(x)

So in the first example above, we would have

f(x) = \frac{4}{x}\\\\f( g(x) ) = \frac{4}{g(x)}\\\\f( g(x) ) = \frac{4}{x^2+9}

In that third step, g(x) was replaced with x^2+9 since g(x) = x^2+9.

Similar steps will happen with the second example as well (when g(x) = x^2)

4 0
3 years ago
Cal had $15. He spent $3 on pens and $8 on a book.
nikklg [1K]
He spent $11 dollars minus tax
3 0
3 years ago
Read 2 more answers
The equation 2x^2 + x - 1 = 0 has two solutions. Find an equation of the form ax^2 + bx + c = 0, which solutions....
leonid [27]

Answer:

Step-by-step explanation:

Let the solution to

2x^2 + x -1 =0

x^2+ (1/2)x -(1/2)

are a and b

Hence a + b = -(1/2) ( minus the coefficient of x )

ab = -1/2 (the constant)

A. We want to have an equation where the roots are a +5 and b+5.

Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.

The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.

So the equation is

x^2-(19/2)x + 22 =0

2x^2-19x + 44 =0

B. We want the roots to be 3a and 3b.

Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and

(3a)(3b) = 9(ab) =9(-1/2)=-9/2.

So the equation is

x^2 +(3/2) x -9/2 = 0

2x^2 + 3x -9 =0.

4 0
3 years ago
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