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Dominik [7]
3 years ago
14

Divide 560 in the ratio of 3:4

Mathematics
2 answers:
bonufazy [111]3 years ago
4 0

Answer:

3:4

3/7*560

=240

4/7*560

=320

4:3=240 & 320

Step-by-step explanation:

goldenfox [79]3 years ago
3 0

Answer:

560 multiplied by \frac{3}{4} equals 420

Step-by-step explanation:

560 * \frac{3}{4} = 140 * 3 = 420.

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If Jemma recieved £69 everyday how much money would she get for a year
telo118 [61]

Answer:

£25185

Step-by-step explanation:

It depends on how many days are in your year. If it's a regular year, it's 365 days. If it's a leap year 366 days.

69 x 365 = 25185

69 x 366 = 25254

5 0
3 years ago
Read 2 more answers
1. Blue ribbon = 3/8 ft
4vir4ik [10]

Answer:

6/8

Step-by-step explanation:

3/8 + 3/8 =  6/8

Hope this helps :D

4 0
2 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
-2 1/3 - (-5) = ? can someone please help me ty!
Eddi Din [679]

Answer: 8/3

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Can you help me pretty please
Elanso [62]

Answer:

A= 13  B=13.1

Step-by-step explanation:

A-B= -0.1

B-A= 0.1

7 0
3 years ago
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