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ZanzabumX [31]
2 years ago
13

How do you solve -5 ( -6 - 3x) + 8 = 83

Mathematics
1 answer:
Kisachek [45]2 years ago
3 0

Answer:

x=3

Step-by-step explanation:

-5(-6-3x)+8=83

Distribute the -5 to both terms in the parentheses by multiplication, multiply the -5 by the -6 and the -3x.

30+15x+8=83

Add like terms, add the 30 and the 8.

38+15x=83

Subtract 38 from both sides to isolate the x on the left side.

15x=45

Divide by 15 on both sides to solve for x.

x=3

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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 curve in polar coordinates is given by: r=9+3cosθ. Point P is at θ=21π/18 .?
ikadub [295]

Answer:

Step-by-step explanation:

Given that a curve in polar coordinates is given by:

r=9+3cosθ

a) At point P, we have

θ=\frac{21\pi}{18}

Substitute to get

r=9+cos \frac{21\pi}{18}\\=9.3932

b) Cartesian coordinate is

x= rcos \theta=3.6934\\y =r sin \tjeta =8.6365

c) At the origin r =0

when r =0

we have

9+3cos\theta=0\\cos\theta =-3

Since cos cannot take values as -3 it doe snot pass through origin.

6 0
3 years ago
Maria and Katy each have a piece of string. When they put the two pieces of string together end to end, the total length is 84in
Serga [27]

Answer: the length of Maria's piece of string is 45 inches.

the length of Katy's piece of string is 39 inches

Step-by-step explanation:

Let x represent the length of Maria's piece of string.

Let y represent the length of Katy's piece of string.

When they put the two pieces of string together end to end, the total length is 84inches. This means that

x + y = 84 - - - - - - - - - - - -1

Maria's string is 6 inches longer than Katy's. This means that

x = y + 6

Substituting x = y + 6 into equation 1, it becomes

y + 6 + y = 84

2y + 6 = 84

2y = 84 - 6 = 78

y = 78/2 = 39

x = y + 6 = 39 + 6

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8 0
3 years ago
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gayaneshka [121]

Answer:

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3 0
3 years ago
Find mABC<br><br> A-54<br> B-135<br> C-15<br> D-126
bagirrra123 [75]
It’s B



Glad I could help
7 0
3 years ago
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