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Thanks,  

It is important that you study what I have done and fully understand it.

Even if you choose to do it a  different way you should be able to be certain that your (and my) methods are correct and you understand why.

"If you say, 'I don't know if my answer right'    then you do not understand well enough.  

Wow you're amazing! Thank you so much for that response. Didn't expect you to go into that much depth. Definitely the best experience I've had on this site by far. 

This problem has already been posted at https://web2.0calc.com/questions/math_23541

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There are different ways of do questions like this. 

$2m\div 5+3=7(4m-3)\\ \frac{2m}{5}+3=7(4m-3)\\ $

Expand the brackets

$\frac{2m}{5}+3=28m-21\\ \text{Multiply both sides by 5 to get rid of the fraction}\\ 5\left(\frac{2m}{5}+3\right)=5(28m-21)\\ \text{simplify both sides seperately}\\ 5*\frac{2m}{5}+5*3=5*28m-5*21\\ 2m+15=140m-105\\ $

NOW the idea is to get all the lots of letters all on the same side, you have  2m on one sides and 140m on the other side.

I want to get rid off the 2m of the left side SO I have TAKE IT AWAY 

but what i do to one side I must do to the other sides as well because the equation must sty balanced!

$2m+15-2m=140m-105-2m\\ simplify\\ 15=138m-105\\ $

Now I neede to get all the numbers without letters onto the same side.

So I will add 105 to both sides.

$15+105=138m-105+105\\ simplify\\ 120=138m$

Only now do I want to seperate the 138 from the m

138m = 138*m

the opposite of times is divide

so   138m divided by 138 = m

also divide is the same a fraction line

so

$138m\div138 = \frac{138m}{138}=m$

SO I need to divide both sides by 138

$\frac{120}{138}=\frac{138m}{138}\\ \frac{120}{138}=m\\ m=\frac{120}{138}\\ m=\frac{20}{23}$

It is better to keep it as a fraction.

So would m=0.8695...?

I did some stuff differently to get that answer because of what I've been taught but I'm probably wrong.

You said to subtract 2m from both sides but wouldn't you be dividing since 2m is multiplication? And when it comes to doing it to both sides which number on each side would you do it to? All of them or just one? So in 2m÷5+3=28m-21 would you multiply 3, 28m and -21 or just 28m or -21?

Thank you again!

The probability of drawing two black cards without replacement is 23/94 , and the probability of drawing one black card is 1/2 . What is the probability of drawing a second black card, given that the first card is black?

The prob of drawing a black card is a half so half the cards are black

Prob of drawing two balck cards = prob of drawing one * prob of drawing a second one = 23/94

so

$\frac{1}{2}*P(second\; black) = \frac{23}{94}\\ P(second\; black) = \frac{23}{94}*2\\ P(second\; black) = \frac{23}{47}\\$

2m÷5+3=7(4m-3)

$2m\div 5+3=7(4m-3)\\ \frac{2m}{5}+3=7(4m-3)\\$

Here is a start.

1) Expand the brackets on the RHS

2) Multiply both sides by 5. that will get rid of the fraction.

3) Subtract 2m from BOTH sides

then see if you can work out how to finish.

Remember that if you do something to one side you MUST do it to the other side as well

The equation must stay balanced.

Give that a go and let us know how you get on. 

Simplify the following:
2 - (x + 1)/(x - 2) - (x - 4)/(x + 2)

Put each term in 2 - (x + 1)/(x - 2) - (x - 4)/(x + 2) over the common denominator (x - 2) (x + 2): 2 - (x + 1)/(x - 2) - (x - 4)/(x + 2) = (2 (x - 2) (x + 2))/((x - 2) (x + 2)) + ((-x - 1) (x + 2))/((x - 2) (x + 2)) + ((4 - x) (x - 2))/((x - 2) (x + 2)):
(2 (x - 2) (x + 2))/((x - 2) (x + 2)) + ((-x - 1) (x + 2))/((x - 2) (x + 2)) + ((4 - x) (x - 2))/((x - 2) (x + 2))

(2 (x - 2) (x + 2))/((x - 2) (x + 2)) + ((-x - 1) (x + 2))/((x - 2) (x + 2)) + ((4 - x) (x - 2))/((x - 2) (x + 2)) = (2 (x - 2) (x + 2) + (-x - 1) (x + 2) + (4 - x) (x - 2))/((x - 2) (x + 2)):

(2 (x - 2) (x + 2) + (-x - 1) (x + 2) + (4 - x) (x - 2))/((x - 2) (x + 2))

(x - 2) (x + 2) = (x) (x) + (x) (2) + (-2) (x) + (-2) (2) = x^2 + 2 x - 2 x - 4 = x^2 - 4:
(2 x^2 - 4 + (-x - 1) (x + 2) + (4 - x) (x - 2))/((x - 2) (x + 2))

2 (x^2 - 4) = 2 x^2 - 8:

(2 x^2 - 8 + (-x - 1) (x + 2) + (4 - x) (x - 2))/((x - 2) (x + 2))

(x + 2) (-x - 1) = (x) (-x) + (x) (-1) + (2) (-x) + (2) (-1) = -x^2 - x - 2 x - 2 = -x^2 - 3 x - 2:
(-8 + 2 x^2 + -x^2 - 3 x - 2 + (4 - x) (x - 2))/((x - 2) (x + 2))

(x - 2) (4 - x) = (x) (4) + (x) (-x) + (-2) (4) + (-2) (-x) = 4 x - x^2 - 8 + 2 x = -x^2 + 6 x - 8:
(-x^2 + 6 x - 8 + 2 x^2 - x^2 - 3 x - 8 - 2)/((x - 2) (x + 2))

Grouping like terms, 2 x^2 - x^2 - x^2 + 6 x - 3 x - 8 - 8 - 2 = (-3 x + 6 x) + (-8 - 2 - 8) + (2 x^2 - x^2 - x^2):
((-3 x + 6 x) + (-8 - 2 - 8) + (2 x^2 - x^2 - x^2))/((x - 2) (x + 2))

6 x - 3 x = 3 x:

(3 x + (-8 - 2 - 8) + (2 x^2 - x^2 - x^2))/((x - 2) (x + 2))

2 x^2 + (-x^2 - x^2) = 0:

(3 x + (-8 - 2 - 8))/((x - 2) (x + 2))

-8 - 2 - 8 = -18:

(3 x + -18)/((x - 2) (x + 2))

Factor 3 out of 3 x - 18:

(3 (x - 6))/((x - 2) (x + 2))=3x - 18 / x^2 - 4

a = 3      and     b=18

$\begin{pmatrix} n\\ a \end{pmatrix} = \frac{n!}{(a)!(n-a)!}$

substitute  n-1  for  n  and simplify.   Then you will have your answer.

Give it a go and get back with your answer or your difficulty.  

Please no one answer over the top of me, wait for guest to give a proper response.  Thanks.

Thanks for your time

Like what guest had said, drop two perpendicular heights from the left and right to the bottom of the trapezoid. Now, we have two 30-60-90 triangles and using those properties, the angle opposite to 90 degrees is two, so the angle opposite to 30 degrees is 1. Adding up all the side lengths, the perimeter is thus $2+2+2+1+2+1=2(4)+2=\boxed{10}.$

Drop a perpendicular from the top left corner to the bottom.  Use sine of 30^o (or cosine of 60^o) to determine the length of the base of the triangle that you formed.  It'll be the same when you do it to the right hand side.  Add up all the lengths. 

Use pythagorean theorem to find the side of the smaller triangle. 

$1^2+b^2=5^2$

$b=2\sqrt6$

So continuing with the pythagorean thereom $2\sqrt{6}^2+5^2=c^2$

$c=\sqrt{37}$

I am assuming the problem asks for a square root version.

Yes I now realize my mistake. If you look in the beginning of my answer you see that I used $5$ instead of  $5^2$

Here is my edited answer...

Use pythagorean theorem to find the side of the smaller triangle. 

$1^2=b^2=5^2$

$1+b^2=25$

$b^2=24$

$b=\sqrt{24}$

So continuing with the pythagorean thereom $\sqrt{24}^2+5^2=c^2$

$24+25=c^2$

$24+25=49$

$\sqrt{49}=7$

Therefore the answer is $7$

Thanks, ElectricPalov and SoulSlayer615 for correcting my previous error,

$\pi$

$\text{I assume this is supposed to be}\\ \sum \limits_{k=0}^{600}i^k$

The sum of any 4 adjacent terms of this series is 0.

600 is divisible by 4.

There are 601 terms, 600 of which will sum to zero leaving the last (or first) term i⁶⁰⁰ = i⁰ = 1

A convex quadrilateral ABCD is cyclic if and only if its opposite angles are supplementary.

Seeing as how A and C are both acute angles it's not possible that they are supplementary.

Thus ABCD is not a cyclic quadrilateral.

Use pythagorean theorem to find the third side of the SMALLER triangle first.....

    Then use the pythagorean theorem to find the hypotenuse of the LARGER triangle

        Let me know what you find......

Well...I came up with a different answer than KeyLimePi......

Smaller triangle side.....

   1^2 +s^2 = 25

    s=sqrt24

Then using the Pythag theorem on the larger triangle:   sqrt(24)^2 + 5^2 = Hyp^2

                                                                                              24 +        25 = hyp^2

                                                                                                      49= hyp^2

                                                                                                        7=? on the larger triangle

The denominator cannot = 0   This would be 'non-permissable'

16-16a^2   cannot = 0

   Let's find where that is:

       16-16a^2 = 0

         16 = 16a^2

            1= a^2

                a = 1 or -1    <-------Non-permissable points

Thank you for answering my question. However, your answer is not in the choices. So what I did is that I divided 23/94 by 1/2 like 23/1 and 94/2 and I got 23/47 which I think is right. Thank you for answering tho!

I'm going to go out on a limb here.  I don't know formal methods of probability and statistics.  But, I think as follows:

I'm assuming that we're using a standard deck of playing cards, without the jokers.  52 cards.   

I understand why the probablilty of drawing 1 black card is 1/2.  Half the cards are black, and the other half are red. 

26/52 = 1/2    that's simple enough.

Now then, about drawing the next card.

What happened with the first card is independent of the second card.  We aren't asking about two draws, only one draw.

After drawing a black card earlier, that leaves 25 black cards in the deck, and a total of 51 cards.

I say the chance of drawing a black card the second time is 25/51. 

One of my teachers once told us that if you flip a coin and get 10 tails in a row, the odds of getting a tail of the 11th toss are still 1/2.  What went before has no bearing on what comes next.  That is not the same as asking the odds of flipping 11 tails in a row... which would be astronomical. 

.

Thanks Tiggsy.    

I shall learn from your answer properly in the morning :)