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igomit [66]
2 years ago
11

Write two fractions multiplication expressions that are equivalent to the expression 3/4X2/3

Mathematics
1 answer:
kolbaska11 [484]2 years ago
7 0

Answer:

1st fraction =

2/6*3/2

2nd fraction =

3/6*2/2

Step-by-step explanation:

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PLEASE HELP ASAP!!<br> Find the area of the figure. Round your answer to the nearest hundredth.
Ksivusya [100]

Answer:

4 in.

Step-by-step explanation:

length times width = area

actually im not really sure lol. we havent done this in a while

8 0
2 years ago
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This is Algebra btw.
Brums [2.3K]
<h2>Step-by-step explanation:</h2>

Given equations;

y₁ = 3x - 8               -------------------(i)

y₂ = 0.5x + 7          --------------------(ii)

To fill the table, substitute the values of x into equations (i) and (ii)

=> At x = 0

y₁ = 3(0) - 8 = -8

y₂ = 0.5(0) + 7 = 7

=> At x = 1

y₁ = 3(1) - 8 = -5

y₂ = 0.5(1) + 7 = 7.5

=> At x = 2

y₁ = 3(2) - 8 = -2

y₂ = 0.5(2) + 7 = 8

=> At x = 3

y₁ = 3(3) - 8 = 1

y₂ = 0.5(3) + 7 = 8.5

=> At x = 4

y₁ = 3(4) - 8 = 4

y₂ = 0.5(4) + 7 = 9

=> At x = 5

y₁ = 3(5) - 8 = 7

y₂ = 0.5(5) + 7 = 9.5

=> At x = 6

y₁ = 3(6) - 8 = 10

y₂ = 0.5(6) + 7 = 10

=> At x = 7

y₁ = 3(7) - 8 = 13

y₂ = 0.5(7) + 7 = 10.5

=> At x = 8

y₁ = 3(8) - 8 = 16

y₂ = 0.5(8) + 7 = 11

=> At x = 9

y₁ = 3(9) - 8 = 19

y₂ = 0.5(9) + 7 = 11.5

=> At x = 10

y₁ = 3(10) - 8 = 22

y₂ = 0.5(10) + 7 = 12

The complete table is attached to this response.

(ii) To find the solution of the system of equations using the table, we find the value of x for which y₁ and y₂ are the same.

As shown in the table, that value of <em>x = 6</em>. At this value of x, the values of y₁ and y₂ are both 10.

5 0
2 years ago
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 />
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2 years ago
Look at bth plz help tysm
tangare [24]

Answer:

d

Step-by-step explanation:

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2 years ago
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У = х — 22<br> Perpendicular?
Romashka-Z-Leto [24]

Answer:

i got y=-x

Step-by-step explanation:

c

7 0
3 years ago
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