Point Q'Q

Mathematics

2 answers:

LekaFEV [45]3 years ago

5 0

Ummmmmmmmmmmmmmmmmmmmm

Stells [14]3 years ago

4 0

Answer:

9,2

Step-by-step explanation:

i got it on khan

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Mekhanik [1.2K]

Answer:

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Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Let c be the curve of intersection of the parabolic cylinder x2 = 2y, and the surface 3z = xy. find the exact length of c from t

Mandarinka [93]

Parameterize the intersection by setting $x(t)=t$ , so that

$x^2=2y\iff y=\dfrac{x^2}2\implies y(t)=\dfrac{t^2}2$
$3z=xy\iff z=\dfrac{xy}3\implies z(t)=\dfrac{t^3}6$

The length of the path $C$ is then given by the line integral along $C$ ,

$\displaystyle\int_C\mathrm dS$

where $\mathrm dS=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dz}{\mathrm dt}\right)^2}\,\mathrm dt$ . We have

$\dfrac{\mathrm dx}{\mathrm dt}=1$
$\dfrac{\mathrm dy}{\mathrm dt}=t$
$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{t^2}2$

and so the line integral is

$\displaystyle\int_{t=0}^{t=2}\sqrt{1^2+t^2+\dfrac{t^4}4}\,\mathrm dt$

This result is fortuitous, since we can write

$1+t^2+\dfrac{t^4}4=\dfrac14(t^4+4t^2+4)=\dfrac{(t^2+2)^2}4=\left(\dfrac{t^2+2}2\right)^2$

and so the integral reduces to

$\displaystyle\int_{t=0}^{t=2}\frac{t^2+2}2\,\mathrm dt=\dfrac{10}3$

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3 years ago

Four is no less then the quotient of a number x and 2.1

Alecsey [184]

X/2.1<4 X<4×2.1 X<8.4

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4 years ago

How many tens are in 54

mestny [16]

Five tens are in 54, with 4 ones left over. hope this helped!!

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3 years ago