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4 years ago

Find the center and radius of the circle (x+3)^2+(y-1)^2=81

Mathematics

1 answer:

sasho [114]4 years ago

5 0

The equation of a circle is written as ( x-h)^2 + (y-k)^2 = r^2

h and k is the center point of the circle and r is the radius.

In the given equation (x+3)^2 + (y-1)^2 = 81

h = -3

k = 1

r^2 = 81

Take the square root of both sides:

r = 9

The center is (-3,1) and the radius is 9

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Factor factorization of 9

Wittaler [7]

Answer: 3x3

Step-by-step explanation: 3 is a prime number

5 0

3 years ago

How can I solve this problem?

Rus_ich [418]

(f+g)(x)=0
f (x) + g (x)=0

now,
substitute their values

x^2 -4x + x -18 = 0
x^2 -3x -18 =0
x^2 - 6x + 3x - 18 =0
x (x-6) + 3 (x-6) = 0
(x-6)(x+3) = 0

we have to find it's zeroes (or roots) now:

x-6 =0
x = 6

x +3 =0
x = -3

7 0

4 years ago

= 1 square foot Find the area of the figure. square feet

Alborosie

Is there more to this question?

6 0

4 years ago

Read 2 more answers

A cookie recipe required 1 1/3 cups of sugar. If Mrs. Lee has 7 cups of sugar, how many recipes of cookies can she bake? Need an

Vitek1552 [10]

$\frac{21}{3} = 7 \: cups$

$\frac{21}{3} ÷ \frac {4}{3} = 5 \frac{1}{3}$

She can make 5 batches of cookies and have

$\frac{1}{3} \: cups$

left

3 0

3 years ago

HALLP QUICKKKK

Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
 $S_{a}=S_{l}-2$
We also know that in the lake the boat sailed for 1 hour longer than it sailed in the river, so:
 $t_{l}=t_{a}+1$

Now, we can set up our equations.
Speed of the boat traveling in the river:
 $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

3 0

4 years ago