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DENIUS [597]
2 years ago
8

Plzzz help meh I don't like this

Mathematics
2 answers:
Darina [25.2K]2 years ago
7 0
16•6 is equal to 96cm
amid [387]2 years ago
3 0

Answer:

Up top it say 6cm

Step-by-step explanation:

But I can’t really see

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Subtract: ( - 2x4 - 4y + 4z² +6) - (-9x4 – 3y + 4z? +9) 0 - 7x² + y + 3 0 - 11x4 - 7 - 3​
Vinil7 [7]

Answer:  

36x-3y+4z^2+26-7x^2

Step-by-step explanation:

Okay well for starters, you can't subtract that because it's an expression, not an equation. And secondly, there was '?' which I'm going to assue=me was a typing error.

Now to actually simplify this mess of an expression, you just need to combine the like-terms! Once you do that you really can't do anything else, since you don't know the varibles.

5 0
2 years ago
What could x be substituted for x 6x + -4 = 2x + 8
Rzqust [24]
6x-4=2x+8
mnus 2x both sides
4x-4=8
add 4 to both sides
4x=12
divide 4 both sides
x=3
6 0
3 years ago
When graphing a system of equations with INFINITE SOLUTIONS,
vaieri [72.5K]

Answer:

Step-by-step explanation:

Hgbuf7g

4 0
2 years ago
Read 2 more answers
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi<span />
4 0
2 years ago
Hunter placed 2 red, 3 blue, 1 yellow, 8 black, and 7 white straws into a bag.
Sphinxa [80]
7/21 or 1/3rd, it's just a ratio of white to all, there's 7 white straws and 21 in all, then you can simplify it down to 1/3rd.
8 0
2 years ago
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