All categories

3 years ago

Complete the square and then find the center and radius from the circle equation

Mathematics

2 answers:

Pavel [41]3 years ago

6 0

Center is 2,-4 the radius is 5

kirza4 [7]3 years ago

4 0

Answer:

center: (2, -4); radius: 5

Step-by-step explanation:

Group x-terms and y-terms. Add the squares of half the coefficient of the linear term in each group. It can be convenient to subtract the constant, too.

(x^2 -4x) +(y^2 +8y) = 5

(x^2 -4x +4) +(y^2 +8x +16) = 5 + 4 + 16

(x -2)^2 +(y +4)^2 = 5^2

Comparing this to the form of a circle centered at (h, k) with radius r, we can find the center and radius.

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

(h, k) = (2, -4) . . . . . the circle center

r = 5 . . . . . . . . . . . . the radius

You might be interested in

The function f(x) = –x2 – 4x + 5 is shown on the graph. On a coordinate plane, a parabola opens down. It goes through (negative

sertanlavr [38]

The true statement about the function f(x) = -x² - 4x + 5 is that:

The range of the function is all real numbers less than or equal to 9.

<h3 /><h3>What is the domain and range for the function of y = f(x)?</h3>

The domain of a function is the set of given values of input for which the function is valid and true.

The range is the dependent variable of a given set of values for which the function is defined.

The domain of the function: f(x) = -x² - 4x + 5 are all real number from -∞ to +∞

For a parabola ax² + bx + c with the vertex $\mathbf{(x_v,y_v)}$

If a < 0, then the range is f(x) ≤ $\mathbf{y_v}$

If a > 0, then the range f(x) ≥ $\mathbf{y_v}$

Here; a = -1,

The vertex for an up-down facing parabola for a function y = ax² + bx + c is:

$\mathbf{x_v = -\dfrac{b}{2a}}$

Thus,

vertex $\mathbf{(x_v,y_v)}$ = (-2, 9)

Range: f(x) ≤ 9

Therefore, we can conclude that the range of the function is all real numbers less than or equal to 9.

Learn more about the domain and range of a function here:

brainly.com/question/26098895

#SPJ1

5 0

2 years ago

Solve 8[7-(6x-6)]+6x=0

bazaltina [42]

8[7 - (6x - 6)] + 6x = 0

8[7 - 6x + 6] + 6x = 0

56 - 48x + 48 + 6x = 0

104 - 42x = 0 .

- 42x = - 104 .

x = 52 / 21

x ≈ 2.476

5 0

3 years ago

Read 2 more answers

The Math 123 book includes about 230 pages (not including the back parts of the book). If you aim to complete 65% of all the pag

alina1380 [7]

Answer: 150 pages

Step-by-step explanation:

I usually just multiply the number by the percentage and it works out just fine :)

8 0

3 years ago

WILL GIVE BRAINLIEST. Joanna saves nickels and quarters in a jar. If she saved five times as many nickels as quarters, and her c

vekshin1

Answer:

78 quarters

Step-by-step explanation:

I think because, I created an equation, 5x = 390 and solves to get 78

3 0

3 years ago

For a quality control test, a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from it

amm1812

Answer:

95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

Step-by-step explanation:

We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.

The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average mpg = 15.6 mpg

s = sample standard deviation = 1.9 mpg

n = sample of minivans = 100

$\mu$ = population average mpg

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.987 < $t_9_9$ < 1.987) = 0.95 {As the critical value of t at 99 degree

of freedom are -1.987 & 1.987 with P = 2.5%}

P(-1.987 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.987) = 0.95

P( $-1.987 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.6-1.987 \times {\frac{1.9}{\sqrt{100} } }$ , $15.6+1.987 \times {\frac{1.9}{\sqrt{100} } }$ ]

= [15.22 mpg , 15.98 mpg]

Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.

7 0

3 years ago