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3 years ago

Find the center and radius of x^2+y^2=9

2 answers:

7 0

$(x-x_1)^2+(y-y_1)^2=r^2\\
\text{center}=(x_1,y_1)\\\\
x^2+y^2=9\\
(x-0)^2+(y-0)^2=9\\
\text{center}=(0,0)\\
r=\sqrt9=3
$

riadik2000 [5.3K]3 years ago

6 0

$x^2+y^2=9\Leftrightarrow (x-0)^2+(y-0)^2=3^2$ hence the center is (0,0) and the radius is 3

Read more about expressions at:

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4 0

1 year ago

There are 12 more girls than boys in a drama club. The ratio of girls to boys is 9:5. How many girls and how many boys are in th

lana [24]

Answer:

27 girls and 15 boys

Step-by-step explanation:

To solve this, I just used guess and check.

9:5 is a ratio, which means we can multiply each side by any number as long as we do it to both sides.

If we multiply the ratio by 3, we get (9 x 3) : (5 x 3) which is 27:15

And as it turns out, 27-15 is 12, which in this context means there are 12 more girls in drama club, so it works out.

7 0

3 years ago

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Select one of the factors of x3y2 8xy2 − 5x2 − 40. (xy2 5) (x2 4) (xy2 − 5) (x2 − 8) 4. select one of the factors of 5x2 7x 2. (

umka2103 [35]

The correct answers are:

(xy² − 5); none of the above; and (4x − 3)

Explanation:

To factor the first one, we will use factoring by grouping. First group the first two together and the second two together:

(x³y²+8xy²)+(-5x²-40)

Find the GCF of each group. The GCF of the first one is xy² and the GCF of the second is -5. Factor these out of each group:

xy²(x²+8)-5(x²+8)

We now have a GCF of the terms as we have rewritten them. Factoring out (x²+8), we have:

(x²+8)(xy²-5)

To factor the second question, we want factors of 10 that sum to 7. 5 and 2 work for this, so this is how we will split up the x term in our polynomial:

5x²+5x+2x+2

Group together the first two and the last two:

(5x²+5x)+(2x+2)

Find the GCF of each group. The first one has a GCF of 5x and the second has a GCF of 2. Factor these out:

5x(x+1)+2(x+1)

The GCF of the rewritten terms is (x+1). Factoring this out, we have

(x+1)(5x+2)

To factor the third question, we want factors of -24 that sum to 5. -3 and 8 will do this, and this is how we will split up the x term:

4x²+8x-3x-6

Group together the first two and the last two:

(4x²+8x)+(-3x-6)

Find the GCF of each group. The GCF of the first group is 4x and the GCF of the second group is -3. Factor these out:

4x(x+2)-3(x+2)

The GCF of the rewritten polynomial is (x+2). Factor this out:

(x+2)(4x-3)

4 0

3 years ago

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If one day is 20 minutes then how many days are in 10 years

Tanzania [10]

Answer:

73050

Step-by-step explanation:

20*3652.5

8 0

2 years ago

Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

LiRa [457]

we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

Let's verify each case to determine the solution to the problem

case A) $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> the term is not a perfect cube

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

therefore

The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

case B) $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> the term is not a perfect cube

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

case C) $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> the term is not a perfect cube

$a^{22}$ ------> the term is not a perfect cube

$b^{16}$ ------> the term is not a perfect cube

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

case D) $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> the term is not a perfect cube

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

case E) $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

5 0

3 years ago

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