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cluponka [151]
3 years ago
8

Tim estimates that 60÷5.7 is about 10.Will the actual quotient be greater than 10 explain

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0
Well first you can do the problem. So take 60 divided by 5.7. When it is done it will come out to about 10.5263 which is a irrational number. So yes the quotient will be larger than 10.
Hope I helped!
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1/3t + 3/4 = 2/4 -t solve
olga55 [171]
Okay so I am going to summarize the work out process because its a lot to

Here we go

1/3 (t) + 3/4 - 2/4 - t = ?

1/2 (simplify )

(1/3 (T)+3/4 - 1/2 - (t) = ?

t (2) / 2

1 - 2(t) / 2 = ?

3/4 (simplify this )

1/3(t)+ 3/4 - [1 - 2(t) / 2 = ?

1/3 (this is re last one you have to simplify)

L (Denominator): 3

R (Denominator): 4

L: [L.C.M] : 4

R: [L.C.M] : 3

Basically , we just switched the dominators around

So, Therefore The of t is -3/16


T = -3/16
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3 years ago
HELP ASAP!!! QUESTION IN PHOTO! MARKING BRAINLIEST
Norma-Jean [14]
What the question on
3 0
3 years ago
Bags of sugar come in 3 sizes. Small bag: A 175 g bag costs 45p. Medium bag: A 420 g bag costs £1.13. Large bag: A 1.2 kg bag co
kotegsom [21]

Answer:

a 3.889

b 3.735

c 4.013

Step-by-step explanation:

a. 45p is 0  175/45=3.889

b  420/113= 3.735

c1.2kg= 1200g /299 =4.013

/ means divide

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3 years ago
Please, I need it ASAP!!!! I will give brainliest if correct!!!
LenKa [72]

Answer:

recursive: f(0) = 7; f(n) = f(n-1) -8

explicit: f(n) = 7 -8n

Step-by-step explanation:

The sequence is an arithmetic sequence with first term 7 and common difference -8. Since you're numbering the terms starting with n=0, the generic case will be ...

recursive: f(0) = first term; f(n) = f(n-1) + common difference

explicit: f(n) = first term + n·(common difference)

To get the answer above, fill in the first term and common difference values.

4 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
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