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2 years ago

Are 3/4 and 16/20 equivalent? Why or why not is one fraction smaller/larger than the other? How do you know

Mathematics

1 answer:

Jet001 [13]2 years ago

6 0

Answer:

No they are not equivalent because 3/4 would be equivalent to 6/8, 9/12, 12/16/ 15/20. 16/20 would be the greatest fraction because in its simplest form it would be 4/5.

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Find the length of side x in simplest radical form with a rational denominator.

Hitman42 [59]

Answer:

$x =\sqrt 2$

Step-by-step explanation:

Given is a 30° - 60° - 90° triangle.

side x lie opposite to 30° angle

$\implies x = \frac{1}{2}\times \sqrt 8$

$\implies x = \frac{1}{2}\times 2\sqrt 2$

$\implies x =\sqrt 2$

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1 year ago

TEST: portional Relationships - Part 1 Which table shows a proportional relationship between x and y? will give brainliest

Artemon [7]

Answer:

First table

Step-by-step explanation:

The first table shows the relationship y=1/7x

The second table doesn't show a relationship due to (6,15)

The third table doesn't show a relationship due to (3,5)

The fourth table doesn't show a relationship due to (2,10)

So the first table is correct

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2 years ago

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OverLord2011 [107]

Answer:

a^2 + 2ab + b^2

Answered by Gauthmath

3 0

2 years ago

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Estimate the amount of the tip by rounding the bill to the nearest dollar before calculating.

telo118 [61]

Answer:

$16

Step-by-step explanation:

Round : $80.36 : $80

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Sidenote: I hope this answers your questions!

6 0

2 years ago

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For the funtion f(x)= (x+7)^5, Find f^-1 (x)

quester [9]

Answer:

The inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

Step-by-step explanation:

For the function $f(x)= (x+7)^5$ , Find $f^{-1} (x)$

For finding inverse of x,

First let:

$y=(x+7)^5$

Now replace x with y and y with x

$x=(y+7)^5$

Now, solve for y

Taking 5th square root on both sides

$\sqrt[5]{x}=\sqrt[5]{(y+7)^5}\\\sqrt[5]{x}=y+7\\=> y+7=\sqrt[5]{x}\\y=\sqrt[5]{x}-7$

Now, replace y with $f^{-1} (x)$

$f^{-1} (x)=\sqrt[5]{x}+7$

So, the inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

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2 years ago