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2 years ago

Can u help me answer this? A tree is 8.5 feet tall. It grows 0.75 feet each year.

Mathematics

1 answer:

alexandr1967 [171]2 years ago

5 0

Hmm... 8.5 + 0.75y

0.75y + 8.5

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Simplify the following 35÷32<br> plz help me out!!

storchak [24]

35/32 = 35/32

If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction.

35/32 = 1 3/32

8 0

2 years ago

Read 2 more answers

Javier has a model of a cone with a radius of 3 inches and a height of 10 inches. Javier increased his model’s volume by doublin

statuscvo [17]

Answer:

$240 \pi\ in^{3}$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z------> the scale factor

x-----> the volume of the larger cone

y-----> the volume of the original cone

$z^{3}=\frac{x}{y}$

In this problem we have

$z=2$ -----> scale factor

substitute

$2^{3}=\frac{x}{y}$

$8=\frac{x}{y}$

$x=8y$

That means-----> The volume of the larger cone is 8 times the volume of the original cone

Find the volume of the original cone

$V=\frac{1}{3}\pi r^{2} h$

we have

$r=3\ in$

$h=10\ in$

substitute

$V=\frac{1}{3}\pi (3^{2})(10)=30 \pi\ in^{3}$

therefore

The volume of the larger cone is equal to

$8*30 \pi\ in^{3}=240 \pi\ in^{3}$

3 0

2 years ago

Read 2 more answers

How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.