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Hitman42 [59]
2 years ago
13

Help? I dont remember how to do this!!!!​

Mathematics
2 answers:
kenny6666 [7]2 years ago
7 0

Answer:

Ex: Shown in #12. 12 x 12 x 12 x 12= 12 with the power of 4

13. 5 x 5 x 5 x 5 x 5 x 5 x 5 =5 with the power of 7 (simple)

14. 7 x 10 x 10 x 10=7/10 with the power of 3 or 4 (i think)

15. 16 x 10 x 10 x 10 x 10= 16/10 with the power of 4 or 5 (sorry if im wrong)

16. 25,000,000= 2.5 x 10^y (im pretty sure)

Again sorry if im wrong and have a wonderful day!!!

Vesnalui [34]2 years ago
4 0

Answer:

13. 5^7

14. 7^1000

15. 16^10000

25 x 10 x 10 x 10 x 10 x 10 x 10

Step-by-step explanation:

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Step-by-step explanation:

3 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
Mary invests £120000 in a savings account.the account pays 1.5% compound interest per year work out the value of her investment
iris [78.8K]

Answer:

120,027£

Step-by-step explanation:

A=P(1+r/100)^n

120,000(1+1.5/100)^2

=120,027£

hope it helps

5 0
3 years ago
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Mariulka [41]

Answer:

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Step-by-step explanation:

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Answer: C

Step-by-step explanation:

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