A line joins the points A (-2,-5) and B (4, 13). (a) Calculate the length AB.

Roger Waters can scrape up just enough money to rent a canoe for 1 hour and 30 minutes. If he paddles out on a lake at 4 kph and

exis [7]

Answer:

He will have paddled the canoe for 4 km.

a. The distance paddled out is equal to the distance paddled returning.

Step-by-step explanation:

Let d be the distance that Roger Waters paddled out.

Let t be the time for the travel he paddled out and t' be the time for returning.

Then from the formula $time=\frac{distance}{speed}$ ,

t = $\frac{d}{4}$ and t' = $\frac{d}{2}$

Total time for the journey = 1 hour 30 minutes = 1.5 hours

So, t + t' = 1.5

$\frac{d}{4} +\frac{d}{2} =1.5$

$\frac{3d}{4} =1.5$

3d = 6

d = 2 km

The total distance he will have paddled the canoe = 2d = 2(2) = 4 km.

Clearly, the distance he paddled out = the distance he returned.

Hence, the correct option is A.

5 0

3 years ago

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I need help plzzzzz be a good mon,woman

leva [86]

The graph is symmetric when X=40

The revenue earned when the shovel is sold for $20 is the same when the shovel is sold for $60.

Hope this helps! :)

6 0

3 years ago

What do you need to do to add and subtract decimals?

olga nikolaevna [1]

Answer:

Write down the numbers, one under the other, with the decimal points lined up

Put in zeros so the numbers have the same length

Then add using column addition, remembering to put the decimal point in the answer

Step-by-step explanation:

6 0

3 years ago

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An old bone contains 80% of its original carbon-14. Use the half-life model to find the age of the bone. Find an equation equiva

Ira Lisetskai [31]

Answer:

An old bone contains 80% of its original carbon-14 in 1844.6479 years

Step-by-step explanation:

We know that

half life time of C-14 is 5730 years

so, h=5730

now, we can use formula

$P(t)=P_0(\frac{1}{2})^{\frac{t}{h} }$

we can plug back h

and we get

$P(t)=P_0(\frac{1}{2})^{\frac{t}{5730} }$

An old bone contains 80% of its original carbon-14

so,

P(t)=0.80Po

we can plug it and then we solve for t

$0.80P_0=P_0(\frac{1}{2})^{\frac{t}{5730} }$

$0.80=(\frac{1}{2})^{\frac{t}{5730} }$

$\ln \left(0.8\right)=\ln \left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)$

$t=-\frac{5730\ln \left(0.8\right)}{\ln \left(2\right)}$

$t=1844.6479$

So,

An old bone contains 80% of its original carbon-14 in 1844.6479 years

4 0

3 years ago

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The formula for depreciation is V=P(1-r)t where P is the beginning price of the asset, r is the annual rate of depreciation writ

Bad White [126]

Answer:

The value of the car after 5 years is $ 13,035 to the nearest dollar.

Step-by-step explanation:

The formula for depreciation is

$V=P(1-r)^t$

where $P=24,700\:dollars$ is the beginning price of the asset,

$r=12\%$ is the annual rate of depreciation written as a decimal,

$r=0.12$

$t=5$ is the number of years, and V is the value after t years.

Now substitute all these values into the formula to get,

$V=24,700(1-0.12)^5$

This simplifies to,

$V=24,700(0.88)^5$

$\Rightarrow V=24,700(0.5277319168)$

$\Rightarrow V=13,034.97834$

The value of the car after 5 years is $ 13,035 to the nearest dollar.

8 0

4 years ago