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1 year ago

Which fraction is equivalent

Mathematics

2 answers:

ladessa [460]1 year ago

6 0

C) -3/5

In fractions, it does not matter whether the negative sign is placed in the numerator or denominator since a fraction represents division.
For example, 6 divided by -1, or 6/-1, gets you -6 for an answer. -6 divided by 1, or -6/1, also gets you -6 as an answer.

Alina [70]1 year ago

3 0

The answer would be d

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Can someone please help me ASAP

USPshnik [31]

Answer:

Look down below!

Step-by-step explanation:

Fist, make your inequality:

-x + 5y > -10

5y > x - 10

y = .2x - 2

Now, plug in any value of x to find y!

Ex: x = 5...

y = .2 (5) - 2

y = -1

Your first coordinate could be (5, -1)

Hope this helps!

8 0

2 years ago

Look at the data set below:

Elenna [48]

Median: Sort from least to greatest and find the # in the middle.
Sort: 54, 58, 58, 60, 62.
Number in the middle: 58, so the median is 58.
Mode: 58, because it appears most.
Range: Subtracting bigger number from smaller number.
Finding it: 62-54 = 8
If you want to find the mean, then here:
Mean: Solving average (adding all #s together and dividing by how many numbers are there) 58 + 54 + 62 + 58 + 60 = 292.
Divide by 5 = 58.4
Hope that helps you. Good luck with it.
Follow if you want.

6 0

3 years ago

Read 2 more answers

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

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

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$