X^2 = 25/289 (Solve for X)

Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m

Dovator [93]

Option a: $\frac{m^{5} }{162n}$ is the equivalent expression.

Explanation:

The expression is $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ where $m\neq 0, n\neq 0$

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

$\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }$

Cancelling the like terms, we have,

$\frac{3^{-3}m^{5} n^{-1}}{6 }$

This equation can also be written as,

$\frac{m^{5}}{3^{3}6 n^{1} }$

Multiplying the terms in denominator, we have,

$\frac{m^{5} }{162n}$

Thus, the expression which is equivalent to $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ is $\frac{m^{5} }{162n}$

Hence, Option a is the correct answer.

8 0

3 years ago

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A tile is selected from seven tiles, each labeled with a different letter from the first seven letters of the alphabet.

MArishka [77]

Answer:

Step-by-step explanation:

U= {A,B,C,D,E,F,G}

X = {A, B, C, D}

Y = {A, C, E, F}

a) X or Y = {A, B, C, D} and {A, C, E, F} = {A, B, C, D, E, F}

B) X and Y = {A, B, C, D} or {A, C, E, F} = {A, C}

c) Complement of X = {A,B,C,D,E,F,G} - {A, B, C, D} = {B,D,G}

7 0

3 years ago

Which equation best represents the relationship

natka813 [3]

Answer:

I've never had to deal with this before but I would say C

8 0

2 years ago

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6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately 298.87.

(b) The number of years before the capital stock exceeds $100,000 is approximately 46.15 years.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

2 years ago

Question 1 of

AnnyKZ [126]

Answer:

5L/2 < 100

Step-by-step explanation:

Let's get started!

Let :

⇒ Length = L

⇒ Width = L/4

It mentions the perimeter must be less than 100 inches.

Perimeter Formula :

Perimeter (Rectangle) = 2(Length + Width)

Then, the inequality formed will be :

⇒ 2 (L + L/4) < 100

⇒ 2 (5L/4) < 100

⇒ 10L/4 < 100

⇒ 5L/2 < 100

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2 years ago

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