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Svetach [21]
3 years ago
5

Solve the problem below by finding a common denominator. 5/6-2/5

Mathematics
1 answer:
sergeinik [125]3 years ago
8 0

Answer:

I know how to this in one of two ways and I'm not sure if they're correct but anything to help.

5/6-2/5 would get you 1/3 because if you cross multiply and cross out factors that can go into each you cant pretty much get that answer. ( i think)

or

the common denominator would be 30 multiply 5 by 5 and 2 by 6. 25-12 is 13. How I was taught was find a number that both denominators can go into then multiply by that number. So 6*5= 30 so you would multiply 5 by 5 to get 25.

Hope I explained this well, I don't really think so.

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) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle
Fynjy0 [20]

The surface (call it S) is a triangle with vertices at the points

x=0,y=0\implies z=2\implies(0,0,2)

x=0,y=2\implies z=-2\implies(0,2,-2)

x=2,y=0\implies z=-8\implies(2,0,-8)

Parameterize S by

\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to S to be

\vec s_v\times\vec s_u=(20v,8v,4v)

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv

=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}

8 0
3 years ago
Help me with this math question please I'm giving away brainliests ​
Dimas [21]

Answer:

I believe your answer is the second option. My apologies if it is wrong.

Step-by-step explanation:

5 0
3 years ago
Which word BEST describes 3x, 5y, and z in the expression 3x − 5y + z?
zysi [14]
A. terms is the best way to describe them
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3 years ago
Use the Ratio Test to determine whether the series is convergent or divergent.
natka813 [3]

Answer:

The series is absolutely convergent.

Step-by-step explanation:

By ratio test, we find the limit as n approaches infinity of

|[a_(n+1)]/a_n|

a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)

a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)

[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]

= |-3n³/2(n+1)³|

= 3n³/2(n+1)³

= (3/2)[1/(1 + 1/n)³]

Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity

= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity

= 3/2 × 1

= 3/2

The series is therefore, absolutely convergent, and the limit is 3/2

3 0
2 years ago
Points (5, 3) and (6, 2) lie on line k. Which is the slope of the line that is parallel to k?
Over [174]
Slope of k=(3-2)/5-6
=-1
parallel lines have equal slopes
so slope of line parallel to k=-1
8 0
3 years ago
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