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kati45 [8]
3 years ago
9

Solve 12y^-4z^-2/6z^2y^-1

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
3 0

Answer:

2/y^5

Step-by-step explanation:

not sure but i think its this

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allsm [11]

Answer:

x=5.3125

One solution

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3 years ago
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Solve for x in the equation 2x^2-4x-9=29
Andrews [41]

Answer:

x = 1  ±2sqrt(5)

Step-by-step explanation:

2x^2-4x-9=29

Add 9 to each each side

2x^2-4x-9+9=29+9

2x^2-4x=38

Divide by 2

2/2x^2-4/2x=38/8

x^2 -2x =19

Complete the square

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

x^2 -2x +1 = 19+1

(x-1)^1=2 = 20

Take the square root of each side

sqrt((x-1)^2) = ±sqrt(20)

x-1 =  ±sqrt(20)

Add 1 to each side

x-1+1 = 1  ±sqrt(20)

x = 1 ±sqrt(20)

Simplifying the square root of 20

x = 1  ±sqrt(4)sqrt(5)

x = 1  ±2sqrt(5)

3 0
3 years ago
What will be the index of 2 for its value equal to 1/2​
Art [367]

Answer:

the index of 2 is -1

Step-by-step explanation:

2^{-1}

=\frac{1}{2^1}

=\frac{1}{2}

3 0
3 years ago
2. Saul invested an average of $425 per month since age 30 in various securities for his
BabaBlast [244]

Answer:

$507.30

Step-by-step explanation:

-Given the monthly deposits are $425 and the interest rate is 3.5% for 30 years.

-The amount of the investment after 30 years is calculated as;

A=P(1+i/n)^n, n=time \ in \ months\\\\=425(1+0.035/12)^{30\times 12}\\\\=1212.65

-Assuming Saul started saving at age 20, his investment term will be 40 yrs.

-His investment amount is thus:

A=P(1+i/n)^n, n=time \ in \ months\\\\=425(1+0.035/12)^{40\times 12}\\\\=1719.95

#We subtract to find how much more he would have if he started saving at 20;

=A_{20}-A_{30}\\\\=1719.95-1212.65\\\\=507.30

Hence, Saul would have $507.30 more had he started saving 10 years earlier.

5 0
3 years ago
What factors does Jenna need to model for the sides?
gladu [14]
I don’t see a picture or anything? Can you explain more I don’t know the sides
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3 years ago
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