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3 years ago

14

It takes has a 20 minutes to walk to school . it takes Andre 80% as long to walk to school how long dose it take Andre to walk t

o school

1 answer:

Blizzard [7]3 years ago

7 0

You have to find 80% of 20.
Since 80% is equal to 0.8, you multiply 20 by 0.8 to get 16

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Equation in slope-intercept form:

Jobisdone [24]

Answer:

y = -x + 3

Step-by-step explanation:

Find the slope using the formula [ y2-y1/x2-x1 ]. We can use the points (0, 3) and (3, 0) to solve.

0-3/3-0

-3/3

-1

From the graph, the y-intercept is (0, 3). Input all the data we know into the slope intercept form expression [ y = mx + b ].

y = -x + 3

Best of Luck!

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3 years ago

Identify the recursive formula for the sequence given by the explicit formula An) = 20

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Answer: f(n) = {f(1)=-4

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Or Answer B.

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3 years ago

What is the value of x in the equation 5(3x + 4) = 23?

Dvinal [7]

Answer:

x = 1/5

Step-by-step explanation:

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3 0

3 years ago

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What is the value of x in the equation 4x + 8y = 40, when y = 0.8?

Alona [7]

$4x+8y=40\\\\ \ \ \ \ when\ y=0.8:\\\\4x+8\cdot0.8=40\\4x+6.4=40\ \ \ \ |subtract\ 6.4\\4x=40-6.4\\4x=33.6\ \ \ |:4\\\boxed{x=8.4}$

8 0

3 years ago

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago