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-Dominant- [34]
3 years ago
14

Help also dont mind karl

Mathematics
2 answers:
SpyIntel [72]3 years ago
5 0
1 1/2 x 1 1/4 = 1.875
Dvinal [7]3 years ago
4 0

Answer:

15/8

Step-by-step explanation:

the mixed answer is 1 7/8

but also the answer is 15/8

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Does anyone know how to find the side length of this square.if so then, i will mark brainlist.
kvv77 [185]
Hello!

To find the side length you use the equation

a =  \sqrt{2}  \frac{d}{2}

a is side length
d is diagonal

Put in the number we know

a =  \sqrt{2} *  \frac{27}{2}

Divide

a =  \sqrt{2} * 13.5

Multiply

a = 19.091

The answer is 19.09

Hope this helps!
8 0
3 years ago
I need to find the slope-intercept form of the equation of the line through (-5,-2) and (-3,4)
Vlad [161]

Answer:

y=3x+13

Step-by-step explanation:

8 0
3 years ago
The manager of a store increases the price of a popular product by 5%. Let y be the original price of the product. The new price
Sphinxa [80]

Answer:

you will multiply the new price plus the original price

5 0
2 years ago
How fast is a car going if it travels 15 meters in 5 seconds?​
scoray [572]

The car is traveling 3 meters second

Step-by-step explanation:

you divide 15 by 5 and you get 3

5 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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