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Zepler [3.9K]
3 years ago
10

How do i solve a question like:

Mathematics
1 answer:
Lynna [10]3 years ago
3 0
Let x be the original amount as per the ques tion (7/6) x= 420thus x=420*6/7=360thus the interest earned is 420 -360=60
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Steps to solve Tangent of 5π/4
seraphim [82]

Answer:

1

Step-by-step explanation:

Tangent of [5 pi]/4

The expression above is given in radian; when ever you have something like this- take note, it's in radian.

Pi [in radian] = 180 in degrees

Hence [5 pi]/4 = [5 x 180]/ 4 =5 x 45 = 225

So Tan 225 = Tan [225-180] =Tan 45

and Tan 45 =1

8 0
3 years ago
24 boxes of crayons weigh 235.2 ounces. ABOUT how many ounces is each box of crayons?
erastova [34]

You would solve this by dividing the total number of ounces by the amount of boxes there are. So it would be 235.2/24=9.8.

3 0
3 years ago
Read 2 more answers
I did a test for math and got a 64. I did a retake with the help of this app and look back at the other test and it just got me
s344n2d4d5 [400]

Answer:

Well, at least you improved.  But maybe you should use another source besides brainly for your tests.  Sometimes people put the wrong answers just to get points.

I hope you do better next time!  Good luck!!

5 0
3 years ago
Which expression is equivalent to 8?<br><br><br> -48 ÷ 6<br> 48 ÷ (-6)<br> 48 ÷ 6<br> -48 ÷ (-6)
V125BC [204]
Expression 3, 
because 6 and 8 are factors of 48. Since positive 6 & 8 are the two numbers of choice, the product will be positive. My answer is reasonable because if you were to multiply positive 8 and 6 you will get positive 48.
8 0
3 years ago
Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straig
Brilliant_brown [7]

Stokes' theorem equates the line integral of \vec F along the curve to the surface integral of the curl of \vec F over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it T) by

\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)

\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)

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

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of T is traversed in the counterclockwise direction when viewed from above.

Compute the curl of \vec F:

\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)

Then by Stokes' theorem,

\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S

where

\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS

\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv

The integral thus reduces to

\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}

3 0
3 years ago
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