All Answers 358150 Answers

R = V/I

12 = V/8. So V = 12*8 or V = 96 (volts)

Hence. R = 96/I

.

This is asking for the angle whose cosine is 1.21.  However, cosine only takes values between -1 and +1, so there is no such angle.

P( get exactly one contract) = 0.4*0.7+0.6*0.3 = 0.48+0.18 = 0.6

P(not get either) = 0.6*0.7 = 0.42

Not sure about b and c

2sin^2(x)-3cos(x)=0

\(2sin^2(x)-3cos(x)=0\\ 2(1-cos^2(x))-3cos(x)=0\\ 2-2cos^2(x)-3cos(x)=0\\ -2cos^2(x)-3cos(x)+2=0\\ 2cos^2(x)+3cos(x)-2=0\\ Let\;\;y=cos(x)\\ 2y^2+3y-2=0\\ 2y^2+4y-1y-2=0\\ 2y(y+2)-1(y+2)=0\\ (2y-1)(y+2)=0\\ y=\frac{1}{2}\;\;\;\;or\;\;\;\;y=-2\\ cos(x)=\frac{1}{2}\;\;\;\;or\;\;\;\;cos(x)=-2\\ x=2\pi n\pm \frac{\pi}{3}\;\;\;\;or\;\;\;\;no\; solution\\ so\\ x=2\pi n\pm \frac{\pi}{3}\qquad where\;\;n \in Z \)

BERTIE (or anyone who really understands)   Please can you help here:      

 Different (but realted ) Question:

I have six bottomless jars of lollies.    They are  Red, Green, blue, yellow, orange and black respecitvely.

I want to randomly choose 2 lollies. How many combintions are in the sameple space and what is the probability that one is red and one is green.

I am told it is appropriate to use the stars and bars method for a question like this

Thanks very much Alan for the explanation and history of the Monte-Carlo simulation technique.  

Thanks also Nauseated for your simplification explanation and for your reference to that older great post.  :)   

(5ab)(6ab)

\((5ab)(6ab)=30a^2b^2\)

This is garble.

If you want us to answer a question for you then you need to give us something that makes some sense. !

    3a(p-2q)   +2b(2q-p)

=  3a(p-2q)   -2b(-2q+p)

=  3a(p-2q)   -2b(p-2q)

=  (3a-2b)(p-2q) 

If you need me to explain better then ask me to :)

(Try and tell me which bit you do not understand though)

Solution

1

Multiply both sides by 1212

tan40×12=xtan⁡40×12=x

2

Switch sides

x=tan40×12x=tan⁡40×12

Done

Decimal Form: -13.4066

Check Answer ▲

tan40=x12tan⁡40=x12

1

Let x=tan40×12x=tan⁡40×12

tan40=tan40×1212tan⁡40=tan⁡40×1212

2

Cancel 1212

tan40=tan40tan⁡40=tan⁡40

Done

(x^(1/(a-b)*(1/a-c))*(x^(1/b-a)*(1/b-c))*(x^(1/c-a)*(1/c-b))

I rather like that.  You have a real flare for colour  :))

It is a great pic of a clown.   Yes, I definitely like it :)

Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx(x log(x))
The derivative of y is y'(x):
y'(x) = d/dx(x log(x))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = log(x):
y'(x) = log(x) d/dx(x)+x d/dx(log(x))
The derivative of x is 1:
y'(x) = x (d/dx(log(x)))+1 log(x)
The derivative of log(x) is 1/x:
y'(x) = log(x)+1/x x
Simplify the expression:
y'(x) = 1+log(x)
Expand the left hand side:
Answer: |  y'(x) = 1+log(x)

Don't worry Solveit.

Einstein Jr is rude and egotistical beyond belief.  He is also not half as bright as he likes to consider himself.

The words and the picture for this question do not belong together.  The question is nonsense.

The trinangle in the pic with a hypotenuse of 10 has a altitude to the hypotenuse of considerably less than 6.

You have correctly found the area of the biggest triangle in the picture.  (you should have aded cm^2 though)

Heavens only knows what triangle Einstein Jackass wants you to find.

there s a bunch of the way to do that

what do you mean is my answer 37.5 is wrong ? (may be i did a calculation mistake :/  )

That's not right.

Another hint? If you find a value for the area, then you are wrong.

i think:

the hyp of the bigger triangle is devided into two parts  lets say one of them is "m" and the other is 'n'

so we can find m using pytha. theorem:

\(10^2-6^2=8^2\\ m=8\)

and there is a formula that:

\(a^2=m*(m+n)\\ a=10\\ m=8\\ 10^2=8*(8+n)\\ n=\frac92 \)

\(c=m+n\\ c=8+4.5=12.5\)

let s use again pyth. theorem:

\(12.5^2-10^2=7.5^2\\ b=7.5\)

\(A_{triangle}=\frac{7.5*10}{2}=37.5\)

another easier way to do that:

\(h^2=m*n\\ 6^2=8*n\\ n=\frac92\)

Solveit, if you have not done calculus yet your answer was excellent!

Good work Solveit :)