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Subtract 22 from both sides of the equation. x 2 + y 2 + 2 x − 10 y = − 22 Complete the square for x 2 + 2 x . ( x + 1 ) 2 − 1 Substitute ( x + 1 ) 2 − 1 for x 2 + 2 x in the equation x 2 + y 2 + 2 x − 10 y = − 22 . ( x + 1 ) 2 − 1 + y 2 − 10 y = − 22 Move − 1 to the right side of the equation by adding 1 to both sides. ( x + 1 ) 2 + y 2 − 10 y = − 22 + 1 Complete the square for y 2 − 10 y . ( y − 5 ) 2 − 25 Substitute ( y − 5 ) 2 − 25 for y 2 − 10 y in the equation x 2 + y 2 + 2 x − 10 y = − 22 . ( x + 1 ) 2 + ( y − 5 ) 2 − 25 = − 22 + 1 Move − 25 to the right side of the equation by adding 25 to both sides. ( x + 1 ) 2 + ( y − 5 ) 2 = − 22 + 1 + 25 Simplify − 22 + 1 + 25 . ( x + 1 ) 2 + ( y − 5 ) 2 = 4 This is the form of a circle. Use this form to determine the center and radius of the circle. ( x − h ) 2 + ( y − k ) 2 = r 2 Match the values in this circle to those of the standard form. The variable r represents the radius of the circle, h represents the x-offset from the origin, and k represents the y-offset from origin. r = 2 h = − 1 k = 5 The center of the circle is found at ( h , k ) . Center: ( − 1 , 5 ) These values represent the important values for graphing and analyzing a circle.

7 0

2 years ago

A hose fills a hot tub at a rate of 4.89 gallons per minute how many hours will it take to fill a 299 gallon hot tub

Lapatulllka [165]

the answer is 1 hour and 1/4 of a minute

4 0

3 years ago

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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0

2 years ago