Answer:
Let x be the number
double x = 2 x
square the result (2x)^2
(2x)^2 - x subtract the original number
4 x^2 - x = 33
4 x^2 - x - 33 = 0
Solve the quadratic
(1 +- (1 + 4 * 132)^1/2) / 8
= (1 + 528^1/2) / 8 = (1 + 23) / 8 = 3 using the positive exponent
Check:
(2 * 3)^2 - 3 = 36 - 3 = 33
Answer:
16 girls: 33 kids
Step-by-step explanation:
add 16 and 17 then break down the equation and figure out the answer
There are an infinite number of them. The smallest one is 42,
and every multiple of 42 is another one.
I don't remember the official correct way to solve this problem.
I can only show you my own way of doing it.
I take the bigger number, and I go through its multiples, one by one,
until I spot one that's also a multiple of the smaller one.
Example: 6 and 7.
Go through the multiples of 7:
7 ... no, not a multiple of 6
14 ... no, not a multiple of 6
21 ... no, not a multiple of 6
28 ... no, not a multiple of 6
35 ... no, not a multiple of 6
<em><u>42</u></em> ... Yes ! A multiple of 6. yay !
1,000 yes 25% of 4000 is 1000
Answer:
Step-by-step explanation:
1) Use Division Distributive Property: (x/y)^a = x^a/y^a.
2) Multiply both sides by 27^x - 8.
3) Use the product rule: x^a x^b = x^a+b.
4) Simplify 1 + x - 8 to x - 7.
5) Use Definition of Common Logarithm: b^a = x if and only if log<u>b</u><u> </u>(x) = a.
6) Use Change of Base Rule.
7) Use rule of 1: log 1 = 0.
8) Simplify 0/log_27 to 0.
9) Add 7 to both sides.
10) Switch sides.
<u>Therefor</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>7</u><u>.</u>
