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3 years ago

Question 20 please help

Mathematics

1 answer:

Luda [366]3 years ago

5 0

I guess the answer is C or A

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

(b) find expressions for the quantities p2, p3, p4, . . ., and pn representing the amount of atenolol in the body right before t

max2010maxim [7]

<span>Atenolol is a beta adrenergic blocker. This medication acts on these receptors, targeting the contractility and workload of the heart. Therefore, it can be an effective treatment for high blood pressure and angina. In addition to proper dosing, it is also important to monitor heart rate before taking the medication. Dosage ranges may vary, depending on the route and frequency ordered by the provider.</span>

4 0

3 years ago

Please Help Me ASAP<br> Picture Included<br> You Get 15 Points

scZoUnD [109]

I believe the answer is C

7 0

3 years ago

An architect is designing a building each floor will be 12 ft tall. Write an expression for the number of floors the building ca

Svet_ta [14]

Answer:

$\frac{h}{12}$

Where "h" is the height of the building.

Step-by-step explanation:

For this exercise it is important to read and analize carefully the information provided.

According to the data given in the exercise, the height of each floor the arquitect is designing is 12 feet.

You want to know the number of floors of 12 feet tall that building can have for a given building height.

Then, you can let "h" represents the height of the building. This will be the variable in the expression.

In order to find the number of those floors that the building can have for "h", you need divide this height by the height of each floor.

Therefore, you can determine that the expression asked in the exercise is the following:

$\frac{h}{12}$

Where "h" is the height of the building.

8 0

3 years ago

If x-1/x = 9 find the value of x +1/x

iren [92.7K]

Answer:

$\sqrt{85}$

Step-by-step explanation:

Given

x - $\frac{1}{x}$ = 9 ← square both sides

(x - $\frac{1}{x}$ )² = 9²

x² - 2 + $\frac{1}{x^2}$ = 81 ( add 2 to both sides )

x² + $\frac{1}{x^2}$ = 83

Now

(x + $\frac{1}{x}$ )² = x² + $\frac{1}{x^2}$ + 2, thus

x² + $\frac{1}{x^2}$ = 83 + 2 = 85

(x + $\frac{1}{x}$ )²= 85 ( take the square root of both sides)

x + $\frac{1}{x}$ = $\sqrt{85}$

5 0

3 years ago