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3 years ago

The price of apples went from $1.99 per lb to $3.19 per lb in four years. Find the rate of change of the price of apples.

Mathematics

2 answers:

german3 years ago

4 0

Answer: $0.3 per lb per year.

Step-by-step explanation:

We know that the rate of change in price is given by :-

$r=\frac{\text{Change in price}}{\text{Time}}$

Given: The price of apples went from $1.99 per lb to $3.19 per lb in four years.

Change in price = $\$3.19-\$1.99=\$1.2$

Time = 4 years

Now, the rate of change of the price of apples is given by :-

$r=\frac{1.2}{4}=0.3$

Hence, the rate of change of the price of apples= $0.3 per lb per year.

Anna11 [10]3 years ago

3 0

Answer:

$0.30 per lb per year

Step-by-step explanation:

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Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

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Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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3 years ago

The perimeters of the square and the triangle shown below are equal. What is the value of x?

Rudiy27

Perimeter is the sum of all the sides. So we can set up an equation:

$\sf 3x+1+3x+1+3x+1+3x+1=2x+8+x+8+4x+2$

Now solve for 'x', combine like terms:

When it comes to terms with variables it's just like normal addition but we keep the variable:

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So we have:

$\sf 12x+1+1+1+1=7x+8+8+2$

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$\sf 5x=14$

Divide 5 to both sides:

$\boxed{\sf x=2.8}$