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2 years ago

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Q3. a cyclist bikes x distance at 10 miles per hour and returns over the same path at 8 miles per hour. what is the cyclist's av

erage rate for the round trip in miles per hour?

1 answer:

Sladkaya [172]2 years ago

4 0

If he bikes for 10 miles per hour and 8 miles per hour for the same distance x miles, he went 10 miles per hour for x/10 hours, as Distance = Rate*Time and on the way back he would go for x/8 hours. So then he went 2x distance, in x/8 + x/10 hours. Since x/8 + x/10 = 10x/80 + 8x/80 = 18x/80 = 9x/40, he went 2x miles in 9x/40 hours. this can be converted into a rate with the above equation Distance = Rate*Time, so 2x=(9x/40) * Rate, thus we divide by 9x/40 on both sides to get 80x/9x = Rate, the x cancels out, and we get 80/9 Miles per hour.

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A middle school took all of its 6th grade students on a field trip to see a play. The students filled 660 seats, which was 55% o

Elan Coil [88]

Answer:

1200 seats

Step-by-step explanation:

<u> x </u> = <u> 660 </u>

100 55

55x = 66,000

x = 1200

7 0

2 years ago

The oblique rectangular prism below has a length of 6 cm, a width of 8 cm, and a height of 7 cm.What is the area of the base of

Rudiy27

Answer:

Area of the base of the prism = 48cm²

Step-by-step explanation:

A rectangular prism has a rectangular base

Area of a rectangular base, A = length x width

A = 6cm x 8cm = 48cm²

5 0

2 years ago

Read 2 more answers

Troy made a scale drawing of the Statue of Liberty which has an actual height of 305 feet. He decides to use a scale in which 1

sergiy2304 [10]

The height of statue is 12.2 inches in Troy's drawing.

Step-by-step explanation:

Given,

Actual height of statue = 305 feet

Scale used by Troy;

25 feet = 1 inch

1 feet = $\frac{1}{25}\ inches$

305 feet = $\frac{1}{25}*305 \ inches$

305 feet = $\frac{305}{25}\ inches\\$

305 feet = 12.2 inches

The height of statue is 12.2 inches in Troy's drawing.

Keywords: unit rate, division

Learn more about unit rate at:

#LearnwithBrainly

3 0

2 years ago

Kayleigh babysat for 11 hours this week. That was 5 fewer than 2/3 as many hours as she babysat last week, h. Write an equation

lukranit [14]

Answer: The answer is $\textup{x}=\dfrac{2}{3}\textup{h}-5.$

Step-by-step explanation: Given that Kayleigh babysat for 11 hours the present week. Also, this was 5 less than two-third of the number of hours she babysat last week, which is represented by 'h'.

We are to write an equation to represent the number of hours she babysat each week.

So, for that, let 'x' be the number of hours she babysat this week. Then, according to the question, we can write

$\textup{x}=\dfrac{2}{3}\textup{h}-5.$

Also, it is given that

$\textup{x}=11.$

Therefore,

$11=\dfrac{2}{3}\textup{h}-5\\\\\Rightarrow \dfrac{2}{3}\textup{h}=16\\\\\Rightarrow \textup{h}=24.$

Hence, using the above relation, we can find the number oh hours Keyleigh babysat each week.

Thus, the required equation is

$\textup{x}=\dfrac{2}{3}\textup{h}-5,$

where, 'x' and 'h' are the number of hours she sat this week and last week respectively.

4 0

2 years ago

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

2 years ago