Coldplay.....just because YOU can solve it in your head does not mean others can !
Some people need to SEE how to get the answer, need to show their work etc.....
A good portion of the questions asked on this forum can be answered by a good portion of us just by 'doing it in our head'....but we had to learn how to think to be able to do so and this is how we learned when we were starting out...
6/4 = 1.5
1.5*4 = 6
Sure..... fire away !!
It says he HAS 3 !
I think he started with 7, he ate 4 and now he HAS 3 !!!
If jim HAS 3 and THEN eats 4, he has stolen an apple from someone because you can't have negative apples.
So better than mine but if it's so simple why yo need a big fancy equation?
I have a number if I add 6 then divide it by three it makes six what is my number?
(x + 6) /3=6
x+6=18
x=18-6
x=12
3-4=-1
Your description is this:
(x + 6)/3 = 6
Solving for x: Multiply both sides by 3
x + 6 = 18 Subtract 6 from each side
x = 12
Idk, but I think 12
4pq(3-q) + 2p(q-3) - 7q(p - 2pq) - 2p(5q^2 + 3) =
12pq -4pq2 + 2pq - 6p - 7pq + 14pq2 - 10pq2 - 6p group like terms
[12 pq + 2pq - 7pq ] + [ -4pq2 + 14pq2 - 10pq2] + [ -6p - 6p] = terms in the middle set of brackets "cancel"........simplify the terms in the other two sets of brackets.......and we're left with.....
7pq - 12p → two terms
Solve for x: 1-3 sin(x)+2 sin^2(x) = 0 The left hand side factors into a product with two terms: (sin(x)-1) (2 sin(x)-1) = 0 Split into two equations: sin(x)-1 = 0 or 2 sin(x)-1 = 0 Add 1 to both sides: sin(x) = 1 or 2 sin(x)-1 = 0 Take the inverse sine of both sides: x = pi/2+2 pi n_1 for n_1 element Z or 2 sin(x)-1 = 0 Add 1 to both sides: x = pi/2+2 pi n_1 for n_1 element Z or 2 sin(x) = 1 Divide both sides by 2: x = pi/2+2 pi n_1 for n_1 element Z or sin(x) = 1/2 Take the inverse sine of both sides: Answer: | | x = pi/2+2 pi n_1 for n_1 element Z or x = (5 pi)/6+2 pi n_2 for n_2 element Z or x = pi/6+2 pi n_3 for n_3 element Z
you are asked for pints per sundae or pints/sundae
so substitute:
4 pints / 6 sundaes === 4/6 = = 2/3 pints/sundae
I think it IS in fractional notation....
Maybe this?
6 1/3
= 6 x 3/3 + 1/3 = 18/3 + 1/3 = 19/3
-1, 2, 7, 14, 23, 34, 47, 62, 79, 98, 119, 142, 167, 194, 223, 254, 287=a(n) = n^2 - 2.
0.666 recurring pints
Notice the pattern :
-1 2 7 14 23 ???
Difference between terms 3 5 7 9 11
Do you see what the next term might be ???
Simplify the following: 4 p q (3-q)+2 p (q-3)-7 q (p-2 p q)-2 p (5 q^2+3) Factor p out of p-2 p q: 4 p q (3-q)+2 p (q-3)-7 q p (1-2 q)-2 p (5 q^2+3) 4 p q (3-q) = 12 p q-4 p q^2: 12 p q-4 p q^2+2 p (q-3)-7 q p (1-2 q)-2 p (5 q^2+3) 2 p (q-3) = 2 p q-6 p: 12 p q-4 p q^2+2 p q-6 p-7 q p (1-2 q)-2 p (5 q^2+3) -7 p q (1-2 q) = 14 p q^2-7 p q: 12 p q-4 p q^2-6 p+2 p q+14 p q^2-7 p q-2 p (5 q^2+3) -2 p (5 q^2+3) = -6 p-10 p q^2: 12 p q-4 p q^2-6 p+2 p q-7 p q+14 p q^2+-6 p-10 p q^2 Grouping like terms, 12 p q-4 p q^2-6 p+2 p q-7 p q+14 p q^2-6 p-10 p q^2 = (12 p q+2 p q-7 p q)+(-6 p-6 p)+(-4 p q^2+14 p q^2-10 p q^2): (12 p q+2 p q-7 p q)+(-6 p-6 p)+(-4 p q^2+14 p q^2-10 p q^2) 12 (p q)+2 (p q)-7 (p q) = 7 (p q): 7 p q+(-6 p-6 p)+(-4 p q^2+14 p q^2-10 p q^2) -6 p-6 p = -12 p: 7 p q+-12 p+(-4 p q^2+14 p q^2-10 p q^2) -4 (p q^2)+14 (p q^2)-10 (p q^2) = 0: 7 p q-12 p Factor p out of 7 p q-12 p: Answer: | p (7q-12)
-9+-8=-17
It is the Afternoon. Wbu?
(x + 5) (x + 2) = distribute the two terms in the first set of parentheses over the terms in the second set of parentheses......
x(x + 2) + 5(x + 2) =
x^2 + 2x + 5x + 10 = combine like terms
x^2 + 7x + 10
Because it's illegal
How many times have you seen it? It is better the second time
16/(3/4)=12-8^0
=12
2x-1=10
2x=11
x=5.5
.
