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3 years ago

Pls help me find v,w,x,y, and z pls help me I will mark u brainlest

Mathematics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

v 39 degree

z 47 degree

x 94 degree

sry don't know about w and y....

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PLEASE HELP ILL GIVE BRAINLIAST

alexdok [17]

Answer:

Step-by-step explanation:

48-5+ 2x -7 -54 ABE 39. - x=2. 66 - 12 = 54. BC, 15. AB=5 =BC. 6x = 66. - They are. XH. Not. 2) Find the coordinates of the midpoint of the segment with the given endpoints. ... j) Concurrency of Perpendicular Bisectors of a Triangle Theorem: ... Find AE. 40. Find AD. 76. 76. Sy + 20. Find BC. 54. Find AC. 80. Find CD. 76.

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2 years ago

Intel estimates that about 12 quintillion transistors are shipped around the globe each year. If that represents 10,000 times th

ra1l [238]

Answer:

Number on ants on Earth = 1.2 × 10¹⁵

Step-by-step explanation:

1 quintillion = 10¹⁸ or (1000000000000000000)

∴ 12 quintillion = 12 × 10¹⁸

12 quintillion = 10,000 × Number of ants on Earth

Let the number of ants on Earth = n

12 quintillion = 10,000 × n

dividing both sides by 10,000

12 quintillion ÷ 10,000 = n

$n = \frac{12\ \times\ 10^{18} }{10,000} \\n = \frac{12\ \times\ 10^{18} }{10^{4}}\\n = 12\ \times\ \frac{10^{18}}{10^{4}} \\applying\ the\ second\ law\ of\ indices\ (\frac{x^m}{x^n} = x ^{m-n})\\ n = 12\ \times\ 10^{(18 - 4)\\n = 12\ \times 10^{14}\\in\ standard\ form\\n = 1.2\ \times 10\ \times\ 10^{14}\\\therefore n = 1.2\ \times\ 10^{15}$

Number on ants on Earth = 1.2 × 10¹⁵

6 0

3 years ago

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":