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il63 [147K]
3 years ago
8

What is 2/5 + 7/15 ?

Mathematics
1 answer:
Whitepunk [10]3 years ago
3 0

Answer:

<u>13/15 </u>

                               _

Decimal Form: 0.86

Not sure if you would've preferred a step-by-step solution. Sorry! Hope you find this helpful, good luck!

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7x-3=18<br> What is x in the question i cbat figure it out
il63 [147K]
7x-3 = 18
+3 for both sides

7x = 21
devide 7 for both sides

x = 3
3 0
2 years ago
The image of the point (3, -4) under a translation is (0, –5). Find the coordinatesof the image of the point (8,7) under the sam
Sav [38]

We will have the following:

*First: We determine the translation rule:

We have that the point went from (3, -4) to (0, -5), thus the translation rule would be:

(x,y)\to(x-3,x-1)

Second: We apply the translation rule to the other point to determine it's image, that is:

(8,7)\to(8-3,7-1)\to(5,6)

So, the image of the second point is (5, 6).

6 0
1 year ago
What is the measure of the missing side of the right triangle?
finlep [7]
B or 12 would be the correct choice!
Hope this helps and mark as brainliest please!
4 0
3 years ago
Read 2 more answers
Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur
Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

\sigma is parameterized by

\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle

for 0\le u\le7 and 0\le v\le\frac{2\pi}3, and a normal vector to this surface is

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle

with norm

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}

So the integral of f(x,y,z)=x^2+y^2+z^2 is

\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}

6 0
3 years ago
(07.06 LC)
Papessa [141]

It would be the first option

8 0
3 years ago
Read 2 more answers
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