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3 years ago

Help mi plizzzzzzzzzz

Mathematics

2 answers:

mote1985 [20]3 years ago

8 0

With what⁉️⁉️⁉️⁉️⁉️⁉️⁉️⁉️⁉️⁉️⁉️

Nezavi [6.7K]3 years ago

8 0

Answer:

what do you need help with ?

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Is this possible<br> Iiii

Gwar [14]

Yeah Imao you usually do these type of questions on a calculator but I've tried writing it on a calculator and it keeps on saying "error" so I'm gonna guess the question is wrong or sth

but if you type out the question without the bracket before the 12 you'll get 6.63

anyway hope this helps- have a good day bro cya)

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2 years ago

clayton uses 34 of a bottle of hot sauce for every batch of salsa he makes. yesterday, he used 214 bottles of hot sauce. how man

Aleks [24]

If he used 34 bottles for every batch just divide 214 bottles he used by 34 it takes for a batch
he made 6.3 batches

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3 years ago

How do you do ratio charts

Alla [95]

Write several instances of the ratio,then plot the ratios as points on the graph

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3 years ago

Read 2 more answers

Kara was baking a cake. The recipe called for 8.5 cups of flour and she put in 3.2 cups so far. How many more cups of flour does

alina1380 [7]

Answer:

5.3 cups of flour

Step-by-step explanation:

8.5 - 3.2 = 5.3 cups of flour

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3 years ago

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th

jeyben [28]

Consider a circle with radius $r$ centered at some point $(R+r,0)$ on the $x$ -axis. This circle has equation

$(x-(R+r))^2+y^2=r^2$

Revolve the region bounded by this circle across the $y$ -axis to get a torus. Using the shell method, the volume of the resulting torus is

$\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx$

where $2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}$ .

So the volume is

$\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx$

Substitute

$x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt$

and the integral becomes

$\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt$

Notice that $\sin t\cos^2t$ is an odd function, so the integral over $\left[-\frac\pi2,\frac\pi2\right]$ is 0. This leaves us with

$\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt$

Write

$\cos^2t=\dfrac{1+\cos(2t)}2$

so the volume is

$\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}$

6 0

3 years ago