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4 years ago

Carlos purchase in chair for $15 he later saw the chair for $27 to an antique dealer what was the percent markup of the chair

Mathematics

1 answer:

artcher [175]4 years ago

4 0

5inches explanation/ I was beaten by my uncle and his 5 inch hit me in the butt

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This question has three parts. Answer the parts in order.

Whitepunk [10]

Answer:

Step-by-step explanation:

5 0

3 years ago

The average rainfall for each week for the last 4 weeks was 7/12 inch. How much rain fell during the last 4 weeks?

kenny6666 [7]

7/12 times 4 is 2 1/3. So 2 1/3 inches each week

8 0

4 years ago

22. Dante rides his bicycle due west at 10 miles per hour. Annie rides her bicycle due north

Flauer [41]

C in my opinion
#CarryOnLearning

6 0

3 years ago

A right triangle has legs measuring 16 inches and 20 inches what is the length of the hypotenuse

leonid [27]

Answer:

The length of the hypotenuse is $4\sqrt{41}\ in$

Step-by-step explanation:

we know that

In a right triangle

Applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

c is the hypotenuse

a,b are the legs of the right triangle

substitute the values

$c^{2}=16^{2}+20^{2}$

$c^{2}=256+400$

$c^{2}=656$

$c=\sqrt{656}\ in$

Simplify

$c=4\sqrt{41}\ in$

8 0

3 years ago

After the mill in a small town closed down in 1970, the population of that town started decreasing according to the law of expon

wlad13 [49]

Answer:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

Step-by-step explanation:

The natural growth and decay model is given by:

$\frac{dP}{dt}=kP$ (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

$\int \frac{dP}{P} =\int kdt$

$ln|P| =kt +c$

And if we apply exponentials on both sides we got:

$P= e^{kt} e^k$

And we can assume $e^k = P_o$

And we have this model:

$P(t) = P_o e^{kt}$

And for this case we want to find $P_o$

By 1990 we have t=20 years since 1970 and we have this equation:

$143000 = P_o e^{20k}$

And we can solve for $P_o$ like this:

$P_o = \frac{143000}{e^{20k}}$ (1)

By 2019 we have 49 years since 1970 the equation is given by:

$98000 = P_o e^{49k}$ (2)

And replacing $P_o$ from equation (1) we got:

$98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}$

We can divide both sides by 143000 we got:

$\frac{98000}{143000} =0.685 = e^{29k}$

And if we apply ln on both sides we got:

$ln(0.685) = 29k$

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

7 0

3 years ago