A bottle holds 24 ounces of water. It has x ounces of water in it. What does 24 - x represent in the situation?

Mathematics

2 answers:

Misha Larkins [42]3 years ago

4 0

It represents the ounces of water that the bottle can still hold.

fredd [130]3 years ago

4 0

I represents the amount of water the bottle could hold, but does not currently have in it. (Aka the amount of space in the bottle, in terms of oz or water)

I hope this helps!

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The perimeter of the rectangle below is 90 units. find the length of side vy . write your answer without variables.

Salsk061 [2.6K]

The picture in the attached figure

we know that
perimeter of a rectangle=2*[base+height]
base=3y+1
height=2y+3
perimeter=90 units
so
90=2*[(3y+1)+(2y+3)]-----> 90=2*[5y+4]----> 10y+8=90----> y=8.2 units

base=3y+1-----> base=3*8.2+1-----> 25.6 units
height=2y+3----> height= 2*8.2+3---> 19.4 units

the answer is
<span>the length of side vy is (2y+3)-----> 19.4 units</span>

3 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

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

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$