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almond37 [142]
3 years ago
9

Which expression is equivalent to -35 3/5

Mathematics
2 answers:
Arada [10]3 years ago
7 0

Answer: answer is A   -8

Step-by-step explanation:

just took the pretest

Alex787 [66]3 years ago
5 0
You can look at -32 as being -2^5 so if you write (-2^5)^3/5 the 5's cancel out making it -2^3 which equals -8
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What is the perimeter of 6*4
Diano4ka-milaya [45]
24 would be the answer if its wrong i am so sorry but it has to be right

6 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
3 years ago
Find the perimeter of a triangle with side lengths of (3c + 4), (5c - 2), and (4c + 7)
igor_vitrenko [27]

Answer:

12c +9 = Perimeter

Step-by-step explanation:

The perimeter of a triangle is found by adding all the sides

3c+4 + 5c-2 + 4c +7 = Perimeter

Combine like terms

12c +9 = Perimeter

5 0
3 years ago
Is y=4/6-2x a linear expression
weqwewe [10]

Step-by-step explanation:

No it is an equation (equal sign)

specifically y = mx + b form

3 0
3 years ago
Read 2 more answers
I NEED HELP PLZ I WILL GIVE THANKS AND BRAINLIST
Molodets [167]

Answer:

No

Step-by-step explanation:

Because 50+30 is not 180

5 0
3 years ago
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