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4 years ago

Solve for g E = 32 g h

Mathematics

2 answers:

7nadin3 [17]4 years ago

5 0

Answer:

--------- = g

32h

Step-by-step explanation:

We want to isolate g. To accomplish this, divide both sides of the given equation by 32h:

--------- = g

32h

Amiraneli [1.4K]4 years ago

4 0

Answer:

E/(32h) = g

Step-by-step explanation:

E = 32 g h

Divide each side by 32h

E/(32h) = 32 g h/(32h)

E/(32h) = g

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Gauge ran 12 miles in 20 minutes. What is Gaugue's hourly rate?

nikitadnepr [17]

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3 years ago

Read 2 more answers

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

HELP PLZZZZ! thank you

Setler79 [48]

1. Same length.

2.vertically opposite angle.

3.BAC =BDE

4.AC = DE

Hope this helps you ___!!❤

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3 years ago

A dragon can fly 50 feet in 2. Second. About how far can adragon fly in 11 secs

andrey2020 [161]

Answer:

275

Step-by-step explanation:

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3 years ago

Graded Assignment Unit Test, Part 2: Basic Geometric Shapes Answer the questions below. When you are finished, submit this assig

AfilCa [17]

Answer:

In case (a) car makes 105° turn.

In case (b) car makes 75° turn.

In case (c) car makes 105° turn.

Step-by-step explanation:

Figure is redrawn To explain properly (in attachment)

Given : streets are parallel means $\overline{AB}$ ║ $\overline{CD}$ ,

AB - 4th street , CD - 3rd street and XY - King Ave.

∠XLA = 75°

To find : (a) ∠XLB

(b) ∠LMD (left onto 3rd streat means left of car)

∠XLB + ∠XLA = 180° (Linear Pair = 2 adjacent angles are

supplementary)

∠XLB + 75° = 180°

∠XLB = 180 - 75

∠XLB = 105°

∴ In case (a) car makes 105° turn.

∠LMD = ∠XLA = 75° (Corresponding angles of parallel lines are equal)

∠LMD = 75°

∴ In case (b) car makes 75° turn.

∠YMD + ∠LMD = 180° (Linear Pair = 2 adjacent angles are

supplementary)

∠YMD + 75° = 180°

∠YMD = 180 - 75

∠YMD = 105°

∴In case (c) car makes 105° turn.

8 0

4 years ago