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3 years ago

Which set represents the same relation as the table below?

Mathematics

2 answers:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

1) {(0,5), (4,2), (6,9), (9,10)}

Step-by-step explanation:

Edge

san4es73 [151]3 years ago

5 0

9.10 maybe i don’t know honestly its making me answer questions so i tried

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Solve.

Genrish500 [490]

It's B.

4X+2_>-6

-subtract 2 to the other side
4x_>-8

-divide 4 to both sides to get X by itself
X_> -8/4

-Simplify
X_> -2

7 0

3 years ago

3.2 is 20 percent of what number?

ivanzaharov [21]

3.2 is 16% of 20 you turn 20 into a decimal and multiply 3.2 by .20 and get the answer

8 0

4 years ago

Use a surface integral to ﬁnd the surface area of the portion of the sphere xUse a surface integral to ﬁnd the surface area of t

brilliants [131]

Parameterize this surface (call it $S$ ) by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi4$ . The limits on $u$ should be obvious. We find the upper limit for $v$ by solving for $v$ where the sphere and cone intersect:

$\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12$

$\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12$

$\implies\sin^2v=\dfrac12$

$\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4$

Take the normal vector to $S$ to be

$\vec r_u\times\vec r_v=-\cos u\sin^2v\,\vec\imath-\sin u\sin^2v\,\vec\jmath-\cos v\sin v\,\vec k$

(orientation does not matter here)

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv$

$=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}$

4 0

3 years ago

GIVE BRAINLIEST:) please help i dont really understand this?

kolbaska11 [484]

Answer: Answer is B

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500

Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking, $\frac{d(w)}{dt}$ = 3 ft/sec

Rate at which the man is walking, $\frac{d(m)}{dt}$ = 2 ft/sec

Collective rate of both, $\frac{d(m+w)}{dt}$ = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min = 20 min

Now,

Distance traveled by man and woman are $m$ and $w$ ft respectively.

⇒ $m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft$

⇒ $w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft$

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate $\frac{dh}{dt}$ .

Lets calculate h.

Applying Pythagoras formula.

⇒ $h^2=(m+w)^2+500^2$

⇒ $h=\sqrt{(2400+2700)^2+500^2} = 5124.45$

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ $h^2=(m+w)^2+500^2$

⇒ $2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}$

⇒ $\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}$ ...as $\frac{d(500)}{dt}= 0$

⇒ Plugging the values.

⇒ $\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}$ ...as $\frac{d(m+w)}{dt} = 5$ ft/sec

⇒ $\frac{d(h)}{dt} =4.9761$ ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

3 0

4 years ago