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\(17+9=26\)

\(xy=\frac32\qquad\rightarrow\qquad y=\frac{3}{2x} \\~\\ 10x+\frac{3y}{5}\,=\,10x+\frac{3(\frac{3}{2x})}{5}\,=\,10x+\frac{9}{10x}\)

Let's say that this is equal to something,  " n "

n  =  \(10x+\frac{9}{10x}\)

We want to know the minimum value of  n .

There might be a better way to do this, but we can find it by looking at a graph.

Very nice, Mr. Owl  !!!

Wow!!!!....thanks, heureka.....that's almost too easy....!!!

A single day has 86,400 seconds in it. So we divide \(\frac{565,900}{86,400} =\) 6.54976851852 Days.

2x +3y = 9   (1)

x - y = 2     add y to both sides  → x =  2 + y     (2)

Sub ( 2) into ( 1)

2 ( 2 + y)  + 3y  =  9     simplify

4 + 2y + 3y  =  9

5y + 4  =  9       subtract  4 from each side

5y  =  5             divide both sides by 5

y = 1

Abd using (2)  ...   x = 2 + 1  = 3

yes, the answer to 7² is equal to 49 and I'll tell you how.

you have 7² and you convert it into 7*7 and you get 49 which is the answer to 7².

Yes, 49 is the correct answer.

For this specific problem, you are multiplying:   7x7=49

For say \(5^8\) what you do is multiply 5x5x5x5x5x5x5x5= 390,625

The little number shows how many times you multiply and the big number shows what you are multiplying.

sqrt( cos^2(t) + sin^2(t) + (1/cos^2(t))^2 )................(1)

Cos^2(t) + sin^2(t) =1..................................................(2)

(1/cos^2(t))^2 =(sec^2(t))^2=sec^4(t).......................(3)

Just sub (2) and (3) into (1)

Sqrt[1 + sec^4(t)]

7 to the power of 2 is 49. 

Any number to the power of 2 is multiplying that number by itself twice.

So that would be 7*7, which is 49.

So yes, 7^2 is 49.

(Did I mix something up again?)

Just divide twice: 

36 / 108 = 1:3 - or one line for every 3 people.

108/36 =3:1 - or 3 people to every one line.

You could be right. I have a Math Dictionary and gives the following:
1-Combinations with repetitions: Each element can be chosen more than once for a particular combination. The number of possible combinations is then [n + r -1], r.
Example n=4, r=2  a, b, c, d: aa, ab, ac, ad,
                                                           bb, bc, bd,
                                                                   cc, cd,
                                                                        dd,
[4 + 2 -1], 2 =5C2 = 10

Simplify the following:
(-2 x)^(-2)

Multiply each exponent in -2 x by -2:
(1)/((-2)^2 x^2)

(-2)^2 = 4:
 1/(4 x^2)

Solve the following system:
{2 x + 3 y = 9 | (equation 1)
x - y = 2 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:
{2 x + 3 y = 9 | (equation 1)
0 x - (5 y)/2 = (-5)/2 | (equation 2)

Multiply equation 2 by -2/5:
{2 x + 3 y = 9 | (equation 1)
0 x+y = 1 | (equation 2)

Subtract 3 × (equation 2) from equation 1:
{2 x+0 y = 6 | (equation 1)
0 x+y = 1 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 3 | (equation 1)
0 x+y = 1 | (equation 2)
x = 3            and               y = 1

what is 64 to the negetive 73 exponents?

64^-73 =1.4088407314736535309995927066798 x 10^-132

1 centrillion in scientific-notation is 1 multiplied by 10^303

and since you just add zeros when multiplying (for example 100^2=10000) it would be 1 multiplied by 10^606

The ladder, wall and ground form a right-angled triangle so you can use Pythagoras to find the height, h:

h^2 + 1.5^2 = 6.8^2

h = sqrt(6.8^2-1.5^2) m ≈ 6.6 m

General equation for line in slope intercept form is:  y = mx + c  where m is slope and c is y-intercept.

If line is parallel to y = 4x + 7 then it has the same slope, so m = 4

If it passes through (0, 8), the y-intercept is 8  (i.e. it's the value of y when x is 0), so c = 8

Hence y = 4x + 8

. 

With "at least" questions it is often worth thinking about "none" first!

The probability of drawing no Aces from 3 random picks: p_none =  (48/52)*(47/51)*(46/50) →4324/5525

The probability of at least one Ace:   p_{at least one} = 1 - p_none → 1 - 4324/5525 → 1201/5525 ≈ 0.217

.

ok thanks for the input, 

Maybe but it does say '4 different bases' though ://

If they can be repeated then I think it would be  56^4 choices = 56^4 = 9834496 possibilities.

Like so:

.

My understanding is that the 4 bases can be repeated, such as:AAAA, CCCC, GGGG and TTTT. The formula I used is the Binomial expansion with repeats allowed.

These are my answers - accept them at your peril :/

You can ask questions about my logic if you want to. ....

I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if:

(e) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

All in 1                               4ways

2 the same  in 1 box and the odd one in another   4*3=12

2 different  in 1 box and the other in another          4*3=12

all in different boxes          4C2*2 = 6*2=12ways

4+12+12+12 = 40ways

I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if:


(d) The candies are all different and the lunch boxes are all the same?

All in 1                               1

2 in 1 and 1 in another      3

all in different boxes    (one box is empty)   1     

5 ways