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Ray Of Light [21]
3 years ago
15

6x + y = 4.3 +11y Write a formula

Mathematics
2 answers:
a_sh-v [17]3 years ago
5 0
The answer may be 10y = -6x + 4.3 unless they want you to solve it all the way...‍♂️
gogolik [260]3 years ago
4 0

Answer:

6x=4.3+10y

Explanation:

6x+y=4.3+11y

-y -y

6x=4.3+10y

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HELP PLS I WILL MARK BRAINLIEST, DUE TODAYY 100 POINTS!!!!!!!!!
Flauer [41]

Answer:

Y1 intercept = 2

(0,2) (3,-7)

m= (-7 - 2) / (3-0) =-9/3 = - 3

Then y = - 3x + 2

Y2 intercept = - 8

(0,-8) (3,-7)

m = (-7 - - 8) / (3-0) = 1/3

Then y= (1/3)x-8

5 0
3 years ago
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
4 years ago
The angles is 1/12 of the circle. how many degress
Sergio [31]
30 degrees of the circle which is an acute angle.
4 0
3 years ago
A crane lifts a 425 kg steel beam vertically a distance of 66 m. How much work does the crane do on the beam if the beam acceler
Fiesta28 [93]
<span>The solution to the problem is as follows:

W=Fd=mad=425*1.8*66 =50490 J

Therefore, it takes </span><span>50490 J of work the crane will do to accelerate the beam upward at 1.8 m/s^2.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>
6 0
3 years ago
Read 2 more answers
The function f(t)=5tan2t does not have an amplitude and has a period of \pi<br><br> yes or no?
Bess [88]
It's multiplied by 5, so the amplitude is 5.

The period is pi/2 since it's tan(2t) 
8 0
3 years ago
Read 2 more answers
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