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7 days ago

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Which expression is equivalent to (StartFraction 4 m n Over m Superscript negative 2 Baseline n Superscript 6 Baseline EndFracti

on) Superscript negative 2? Assume m not-equals 0, n not-equals 0. StartFraction n Superscript 6 Baseline Over 16 m Superscript 8 Baseline EndFraction StartFraction n Superscript 10 Baseline Over 16 m Superscript 6 Baseline EndFraction StartFraction n Superscript 10 Baseline Over 8 m Superscript 8 Baseline EndFraction StartFraction 4 m cubed Over n Superscript 8 Baseline EndFraction

2 answers:

PIT_PIT [9K]7 days ago

5 0

Question

Which expression is equivalent to $\frac{4mn}{m^{-2}n^6}$ . Assuming $m \neq 0; n\neq 0$

Answer:

$\frac{4m^3}{n^5}$

Step-by-step explanation:

Given

$\frac{4mn}{m^{-2}n^6}$

Required:

Simplify

To achieve simplification, we begin by dividing each function individually

$\frac{4mn}{m^{-2}n^6} = \frac{4m}{m^{-2}} * \frac{n}{n^6}$

According to the properties of indices

$\frac{a^x}{a^y} = a^{x-y}$

Thus, the aforementioned expression can be restated similarly

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{1-(-2)} * n^{1-6}$

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{1+2)} * n^{1-6}$

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{3} * n^{-5}$

From the indices laws

$a^{-x} = \frac{1}{a^x}$

So,

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{3} * \frac{1}{n^5}$

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = \frac{4m^3}{n^5}$

Therefore, $\frac{4mn}{m^{-2}n^6}$ can be regarded as $\frac{4m^3}{n^5}$

Inessa [8.9K]7 days ago

5 0

Answer:

Option D on Edg

Step-by-step explanation:

Just completed the test

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lawyer [9226]

Answer:

Someone has previously provided an answer to this question, and I must attribute it to writersparadise. All credit goes to him/her.

Step-by-step explanation:

There can be various responses to this query. Usually, the next number following 16 has digits that add up to 8.

Explanation:

First number 31 – The digits add up to 3 + 1 = 4

Second number 1031 – The sum of the digits is 1 + 0 + 3 + 1 = 5

Third number 402 – The digits sum to 4 + 0 + 2 = 6

Fourth number 16 – Adding the digits gives 1 + 6 = 7

Thus, the missing number could either be the single digit 8 or any of these two-digit options: 17, 26, 35, 44, 53, 62, 71, or 80. It may also include larger numbers consisting of the digits from the mentioned two-digit figures combined with 0, like 260 or 7001.

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20 days ago

Gus has a fish tank that holds 471047104710 inches^3 3 cubed of water. He is using a cylinder shaped bucket with a radius of 555

tester [8808]

Answer:

Gus must fill the cylindrical bucket three times to get it filled for the tank

Step-by-step explanation:

Extracting essential details from the query:-

*** A fish tank has a volume of 4710 inches^3

*** The tank owner utilizes a cylindrical bucket for filling the tank

*** The bucket has a radius of 5 inches and a height of 20 inches

*** We need to compute how many times Gus has to fill the bucket to fully fill the fish tank.

Since the container used for filling the tank is cylindrical, we will apply the formula for finding a cylinder's volume to derive the bucket's volume used for filling the tank.

Volume of a cylinder = π × r^2 × h. The radius of the cylindrical bucket is stated as 5 inches while the height is mentioned as 20 inches.

The bucket's volume is therefore calculated as 3.14 × 5^2 × 20 = 1570 inches^3

Next, it is given that the fish tank’s capacity is 4710 inches^3. To find out how many times the bucket can fill the tank, we must divide the tank's volume by the bucket's volume:

4710/1570 = 3 full buckets.

Thus, Gus has to fill the bucket three times for filling the fish tank completely without spillage.

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5 days ago

A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds a

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Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

Population mean 'μ' = 10 pounds

Population standard deviation 'σ' = 2.8 pounds

Sample mean 'x⁻' = 9

Significance level ∝ = 0.05

Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

$z= \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$

$z= \frac{9 - 10}{\frac{2.8}{\sqrt{50} } } = \frac{-1}{0.395} = 2.53$

The determined Z value is 2.53

In this case, the critical value Z is 1.96 at the 0.05 significance level

Step(iii):-

Calculated Z value of 2.53 is greater than 1.96 at a significance level of 0.05

Consequently, we reject the null hypothesis H₀

We accept the alternative hypothesis

Conclusion:-

We determine that individuals who sign up for Weight Reducers will experience a weight loss exceeding 10 pounds

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