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sergij07 [2.7K]
2 years ago
15

The width of Maya’s poster is 2 inches shorter than the length. The graph models the possible area (y) of Maya’s poster determin

ed by its length (x).
Which describes what the point (2, 0) represents?
Mathematics
1 answer:
ArbitrLikvidat [17]2 years ago
3 0
200 would be the correct answer
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What is the function rule of (1,30) (2,35) (3,40) (4,45)?
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a&5(a)+25
\end{array}\qquad \qquad y=5x+25

if you have already covered slopes, you could also get it that way, in fact is  simpler that way.
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3 years ago
Find the product of 2 5/6x 1 3/4
Drupady [299]
2 5/6 = 2.83
1 3/4 = 1.75
2.83*1.75= 4.9525
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3 years ago
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The sum of two numbers is 20. When the second number is subtracted from the first number, the difference is 4. Find the two numb
elena-s [515]

Answer:

12 and 8

Step-by-step explanation:

7 0
3 years ago
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Solve this problem on paper using all four steps. A girl scout troop sold cookies. If the girls sold 5 more boxes the second wee
Varvara68 [4.7K]
This is a problem where you should make an equation.
Let's say x is the first week. It would mean
First Week : x
Second Week : x + 5
Third Week : 2(x+5)
You need to add these up to make your equation. Your equation will look something like this :
x+x+5+2(x+5) = 431
Now you simplify.
2x+5+2x+10 = 431
4x+15 = 431
4x = 416
x = 104
That means they sold 104 Boxes in the First Week.
104+5=109
So they sold 109 Boxes the Second Week.
109 x 2 = 218
And they sold 218 Boxes the Third Week.

I hope I was useful!
3 0
3 years ago
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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
3 years ago
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