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ziro4ka [17]
3 years ago
6

What is the product of (3y^-4)(2y^-4)?

Mathematics
2 answers:
borishaifa [10]3 years ago
4 0
(x^a)(x^b)=x^(a+b)

(ab)(cd)=(a)(b)(c)(d)

x^-m=1/(x^m)


(3y^-4)(2y^-4)=
(3)(y^-4)(2)(y^-4)=
(6)(y^-8)=
6/(y^8)
lubasha [3.4K]3 years ago
4 0

Answer:

Product of (3y^{-4})(2y^{-4})=6y^{-8}

Step-by-step explanation:

Given : Expression (3y^{-4})(2y^{-4})

To find : The product of the given expression?

Solution :

(3y^{-4})(2y^{-4})

Applying property of exponent, x^a\times x^b=x^{a+b}

Comparing with given expression x=y , a=-4 and b=-4

=(3)\times(2)\times(y^{-4+(-4)})

Multiply 3 and 2,

=6\times(y^{-8})

=6y^{-8}

Therefore, Product of (3y^{-4})(2y^{-4})=6y^{-8}

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3 years ago
Miguel and three of his friends went to the movies. They originally had a total of $40. Each boy had the same amount of money an
Alenkasestr [34]

Answer:

The money each boy had after buying his ticket is <u>$2.50</u>.

Step-by-step explanation:

Given:

Miguel and three of his friends went to the movies. They originally had a total of $40. Each boy had the same amount of money and spent $7.50 on a ticket.

Now, to find the money each boy had after buying his ticket.

Let the money each boy had after buying his ticket be x.

Total number of boys = 4.

Total money they had = $40.

Money each boy spent = $7.50.

Now, to write an equation:

4(x+7.50)=40

Now, to solve the equation for getting the money each boy had after buying his ticket:

4(x+7.50)=40

4x+30=40

<em>Subtracting both sides by 30 we get:</em>

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<em>Dividing both sides by 4 we get:</em>

x=\$2.50.

Therefore, the money each boy had after buying his ticket is $2.50.

3 0
2 years ago
For what real values of does the quadratic 12x^2+kx+27=0 have nonreal roots
bija089 [108]

Answer:

k<±36

Step-by-step explanation:

Δ<0 (no real roots)

b²-4ac<0

k²-4x12x27<0

k²-1296<0

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Find the 14th term of the geometric sequence 5,-10,20
tamaranim1 [39]

In this specific case, the <em>initial term</em> (a) is 5 and the <em>common ratio</em> (r) is -2

Henceforth, after determining what a and r are, we use the formula for the <em>nth term</em>. Which is:

a_n = a \times r^{n-1}

Therefore, the 14th term is :

\implies 5 \times  { (- 2)}^{14 - 1}  \\\implies 5 \times  {( - 2)}^{13}  \\ =\boxed { -40,960}

Hope it helps!

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77julia77 [94]

Answer:

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Step-by-step explanation:

if you need to round it,

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tenths: 2,934.5

Thousandths: 2,934.467

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