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Romashka-Z-Leto [24]
3 years ago
8

HHEELLPPP PLLLZZZZ RIGTH NOW I'LL MARK U BRAINLIST

Mathematics
2 answers:
netineya [11]3 years ago
7 0

Answer:

A and D I did it on edge:

Ivahew [28]3 years ago
6 0
So the first one is correct and the last option
(0,0), (1,2), (2,4)
and
(0,0), (4,1), (8,2)
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Pls help choose 2 equations !! :) I will mark brainlist
SCORPION-xisa [38]

Answer:

C and E for sure

Step-by-step explanation:

6 0
3 years ago
What is 20% of what number is 96?
malfutka [58]
So first of all you have to change 20% into 0.20 then do 0.20 x 96 and get 19.2 or 19.20 
5 0
3 years ago
Part of a regression output is provided below. Some of the information has been omitted.
quester [9]

Answer:

301.189

Step-by-step explanation:

Given the table :

Source of Variation - - SS - - - df - - - MS - - - - F

Regression - - - - - -3177.17 - - -2 - - 1588.6

Residual - - - - - - - - ______ --17 - - -17.717

Total - - - - - - - - - - 3478.36 - - 19

Calculate the SSR, Sum of Square residual

The Sum of Square RESIDUAL (SSR), Mean Square Residual (MSR) and Degree of Freedom RESIDUAL (DFR) are related by the formular :

MSR = SSR / DFR

Hence,

SSR = MSR × DFR

Fr the table ;

MSR = 17.717 ; DFR = 17

SSR = (17.717 × 17)

SSR = 301.189

5 0
3 years ago
Does this graph show a function? explain how you know
AVprozaik [17]

Answer:

C

Step-by-step explanation:

The vertical line test is basically just drawing a vertical line and seeing if the line intersects the graph more than once. If it does, then it is not a function, if it doesn't than it is a function.

3 0
3 years ago
Read 2 more answers
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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