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MA_775_DIABLO [31]
3 years ago
12

Whats 3.24 X 10^-5 in standard form?

Mathematics
1 answer:
11111nata11111 [884]3 years ago
6 0

Answer:

3.24 × 10 ^ -5

Step-by-step explanation:

3.24 × 10 ^ -5 is already a standard form...

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( 3u + 7 )( 5 - u )= 0​
mixas84 [53]

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The answer is 484. Please let me know if it helped you.

Step-by-step explanation:

5 0
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-x-4(x-2)<br> Help please
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-5x+8

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7 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
An acute angle ? is in a right triangle with cos ? = nine tenths. what is the value of sec ?? square root of nineteen divided by
Anon25 [30]
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6 0
3 years ago
How many triangles would be needed to cover 7/3 trapezoid 7/3 ×3=7 Am I right I think it 21 doe.
jarptica [38.1K]

Answer:

5 or 6 (or 2 or 7)

Step-by-step explanation:

Any quadrilateral can be covered by 2 triangles, so the integer number in 7/3 = 2 1/3 quadrilaerals will require 2×2 = 4 triangles.

The meaning of (1/3) trapezoid determines the number of additional triangles required. If 1/3 trapezoid is a triangle, only one is needed. If 1/3 trapezoid is a quadrilateral, then 2 triangles are needed.

_____

If the 7/3 trapezoids are butted against each other so they make a larger quadrilateral figure, then only 2 triangles are needed.

_____

If each trapezoid is constructed from 3 equilateral triangles, and you want to know the total number of those equilateral triangles in 7/3 such figures, it will be 3 × 7/3 = 7.

The answer depends on problem details not provided here.

7 0
3 years ago
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