All categories

3 years ago

Can someone help me out with this?

Mathematics

1 answer:

dybincka [34]3 years ago

4 0

1/8 + 1/4

the least common multiple of 8 and 4 is 8. Convert 1/8 and 1/4 to fractions with denominator 8.

1/8 + 2/8

Because 1/8 and 2/8 have the same denominator, add them by adding their numerators.

1+2/8

add 1 and 2 to get 3

3/8 or 0.375

You might be interested in

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

###

Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

3 0

3 years ago

A+ 2b = 10 2a + b =6

GenaCL600 [577]

First you can solve for b: Subtract a on both sides and get 2b=-a+10, then divide by 2 to get b by itself and get: b=-1/2a+5

Then you can plug in this equation for b in the other: This would give you: 2a+-1/2a+5=6. Then you can use that equation to solve for a and get: 1.5a+5=6, subtract 5, 1.5a=1, divide by 1.5, a=1/1.5

Then you can plug in the value of a to solve for b.

7 0

2 years ago

Read 2 more answers

A, B and C are points on a circle with a diameter AB

Tamiku [17]

Answer:

36.575

Step-by-step explanation:

you need to find line ab you can do $A^{2} +B^{2} =C^{2}$ for the triangle and get the line you also need to get the squar root of it as it is still squared then devide by 2 to get the rades then times that by pi and there you go you have the area also sorry for spelling I love math not English