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4 years ago

Somebody plz me on this real quick

Mathematics

1 answer:

Kay [80]4 years ago

4 0

I think it would be the second one

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Which set of ordered pairs does not represent a function?

frosja888 [35]

Answer:

(-7,8), (2,1),(-8, -9), (-8,6)

Step-by-step explanation:

the x values repeat

The x values are not supposed to repeat to create a function

7 0

3 years ago

Quiz 2

jek_recluse [69]

There’s a 75% chance that everyone in the group is a girl

5 0

3 years ago

Choose the simplified product.

nikitadnepr [17]

Answer:

I need answer choices to help you hone in on one specific answer but Simplify the radical by breaking the radicand up into a product of known factors

Step-by-step explanation:

4 0

4 years ago

Let F(x, y, z) = (5ex sin(y))i + (5ex cos(y))j + 7z2k. Evaluate the integral C F · ds, where c(t) = 8 t , t3, exp( t ) , 0 ≤ t ≤

maria [59]

I'm going to assume this reads

$\vec F(x,y,z)=5e^x\sin y\,\vec\imath+5e^x\cos y\,\vec\jmath+7z^2\,\vec k$

and the path $C$ is parameterized by

$\vec c(t)=8t\,\vec\imath+t^3\,\vec\jmath+e^t\,\vec k$

with $0\le t\le1$ . Under this parameterization,

$\vec F(x,y,z)=\vec F(x(t),y(t),z(t))=5e^{8t}\sin(t^3)\,\vec\imath+5e^{8t}\cos(t^3)\,\vec\jmath+7e^{2t}\,\vec k$

and

$\mathrm d\vec c=\dfrac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt=(8\,\vec\imath+3t^2\,\vec\jmath+e^t\,\vec k)\,\mathrm dt$

Then in the integral,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec s=\int_0^1\vec F(x(t),y(t),z(t))\cdot\frac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(7e^{3t}+15e^{8t}t^2\cos(t^3)+40e^{8t}\sin(t^3))\,\mathrm dt\approx\boxed{12586.5}$

(It's unlikely that an exact answer can be found in terms of elementary functions)

4 0

3 years ago

Compute <img src="https://tex.z-dn.net/?f=i%2Bi%5E2%2Bi%5E3%2B%5Ccdots%2Bi%5E%7B258%7D%2Bi%5E%7B259%7D" id="TexFormula1" title="

xz_007 [3.2K]

Answer:

-1

Step-by-step explanation:

Note that $i+i^2+i^3+i^4 = i-1-i+1 = 0$ , and this means that every 4 terms, the terms cancel to 0. Therefore, by taking modulo 4, we only need to find $i^{257}+i^{258}+i^{259} = i-1-i = -1$ since $i$ has a cycle of 4.

3 0

3 years ago

Read 2 more answers