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2 years ago

A package of orgainc strwberries costs $1.50 while there in season and $2.25 when not in season what the percent increase

1 answer:

3 0

The percentage increase from $1.50 to $2.25 is 50%.

Steps:
1st- you first need to subtract the difference (increase). You will end up with .75 cents in increase.

2nd- you will then need to divide the increase (.75) by the original number ($1.50). You end up with .50.

3rd- you will then need to multiply the sum (.50) by 100. you end up with 50%

Answer: There is a 50% increase when not in season.

Hope this helps!

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Simplify the expression 2(x - 5)

timofeeve [1]

Answer:

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Step-by-step explanation:

2(x−5)

=(2)(x+−5)

=(2)(x)+(2)(−5)

=2x−10

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2 years ago

The sun came out and the temperature quickly rose 18 degrees. The a thunderstorm hits and the temperature drops 6 degrees. The s

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Just do it backwards.

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2 years ago

On. If your car gets 30 miles per gallon, find the co

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C. $34.50. A car gets 30 miles per gallon, if the gasoline cost $1.15 per gallon, a 900 miles trip will cost $34.50

This exercise can be solved by simply arithmetic.

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3 years ago

Read 2 more answers

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elixir [45]

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2 years ago

Find the mass of the triangular region with vertices (0, 0), (3, 0), and (0, 1), with density function ρ(x,y)=x2+y2.

ololo11 [35]

Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region $\mathcal T$ by computing the double integral of the density function over $\mathcal T$ :

$\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy$

The boundary of $\mathcal T$ is determined by a set of lines in the $x,y$ plane. One way to describe the region $\mathcal T$ is by the set of points,

$\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}$

So the mass is

$\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx$

$=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52$

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2 years ago