All categories

3 years ago

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

1 answer:

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!

You might be interested in

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

4 0

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

4 0

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