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3 years ago

What is the value of y^-5 and x^-3 for x = 2 and y = -4?

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

$a^{-n}=\dfrac{1}{a^n}\\\\x^{-3}=\dfrac{1}{x^3};\ y^{-5}=\dfrac{1}{y^5}\\\\\dfrac{y^{-5}}{x^{-3}}=\dfrac{1}{y^5}:\dfrac{1}{x^3}=\dfrac{1}{y^5}\cdot\dfrac{x^3}{1}=\dfrac{x^3}{y^5}\\\\x=2;\ y=-4=-2^2\\\\\dfrac{x^3}{y^5}=\dfrac{2^3}{(-2^2)^5}=\dfrac{2^3}{-2^{2\cdot5}}=\dfrac{2^3}{-2^{10}}=-\dfrac{1}{2^7}=\boxed{-\dfrac{1}{128}}\to\fbox{C.}$

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An ice-cream factory makes 250 quarts of ice-cream in 2 hrs. how many could be made in 36 hrs

Nutka1998 [239]

Find the unit rate of this scenario. 250/2 = 125 quarts per hour
Now multiply the unit rate by the new amount. 36*125 = 4500

Add the units

4,500 quarts in 36 hours.

8 0

2 years ago

Read 2 more answers

Solve by factoring

True [87]

2×2-13(19÷13)+15=0

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2 years ago

A sock drawer contains eight navy blue socks and five black socks with no other socks. If you reach in the drawer and take two s

Rzqust [24]

Answer:

a. the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b. the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c. the probability of either 2 navy socks is picked or one black & one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Step-by-step explanation:

A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.

If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:

Solution:

total socks = N = 8 + 5 + 0 = 13

a) you will pick a navy sock and a black sock?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick the black sock, total number of socks left is 12.

Let B be the probability of picking a black sock again.

P (B) = 5/12.

Then, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b) the colors of the two socks will match?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick another navy sock, total number of socks left is 12.

Let B be the probability of another navy sock again.

P (B) = 7/12.

Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

Let D be the probability of picking a black socks first.

Then P (D) = 5/13

without replacing the black sock, will pick another black sock, total number of socks left is 12.

Let E be the probability of another black sock again.

P (E) = 4/12.

Then, the probability of picking 2 black sock = P (D & E)

= (5/13 ) * (4/12) = 5/39 = 0.128

Now, the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c) at least one navy sock will be selected?

this means, is either you pick one navy sock and one black or two navy socks.

so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

now either 2 navy socks is picked or one black one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

4 0

2 years ago

A theater is divided into a red section and a blue section. The red section has 350 seats, and the rest of the seats are in the

coldgirl [10]

850.

68,750 = 350(75) + 50x
68, 750 = 26,250 + 50x
-26250
42,500 = 50x / 50
850 = x

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3 years ago

Plss answer!! <br>MERRY Christmas!!! <br>Critical Question. AAAAA

liberstina [14]

Answer:

Step-by-step explanation:

abc = 1

We have to prove that,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

We take left hand side of the given equation and solve it,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$

Since, abc = 1,

$\frac{1}{c}=ab$ and c = $\frac{1}{ab}$

By substituting these values in the expression,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+ab}+\frac{1}{1+\frac{1}{ab}+\frac{1}{a}}$

$=\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{ab+1+b}$

$=\frac{1+b+ab}{1+b+ab}$

$=1$

Which equal to the right hand side of the equation.

Hence, $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

5 0

2 years ago