Answer:
1/10
13/100
4/5
12/25
3/10
63/100
3/5
51/200
2/9
5/11
To prove the last 2 recurring ones:
0.222222... = x
10x = 10 * 0.22222... = 2.222222....
Notice how the decimal part of 10x is the same as for x:
10x - x = 2.2222222... - 0.222222... = 2
10x - x = 9x = 2
x = 2/9
Same procedure for the other one but times by 100 instead:
x = 0.454545...
100x = 45.454545...
100x - x = 45.454545... - 0.454545... = 45
100x - x = 99x = 45
x = 45/99 = 5/11
Answer:
1). 9 red 6 fancy 2). 50 shirts 20 pants 3). 62 purple 31 red 4). 19 dimes 9 quarters
Step-by-step explanation:
Most are systems of equations. So, they will have more than one equation.
1). x (Red wrapping paper) y (fancy print paper)
x + y = 15
2x + 4y = 42
Solved: x = 9 y = 6
2). x (shirts) y (pants)
x + y = 70
12x + 30y = 1200
Solved: x = 50 y = 20
3). x (purple) y (red)
I ended up dividing 93 by 3 and then multiplying it by 2 for purple and keeping the other third for red. Not quite sure how to fit this into an equation, its just simple math.
Solved: x = 62 y = 31
4). x (dimes) y (quarters)
.10x + .25y = 4.15
x = y - 10
Solved: x = 19 y = 9
Answer:
Please use " ^ " for exponentiation: x^2 + 2x + 8 ≤ 0.
Let's solve this by completing the square:
x^2 + 2x + 8 ≤ 0 => x^2 + 2x + 1^2 - 1^2 + 8 ≤ 0. Continuing this rewrite:
(x + 1)^2 + 7 ≤ 0
Taking the sqrt of both sides: (x + 1)^2 = i*sqrt(7)
Then the solutions are x = -1 + i√7 and x = -1 - i√7
There's something really wrong here. I've graphed your function, x^2 + 2x + 8, and can see from the graph that there are no real roots, but only complex roots. Please double-check to ensure that you've copied down this problem correctly.
For a fraction, if the denominator turns to 0, the fraction becomes undefined, and therefore, that's a restriction on a rational.
now, what values of "x" makes the denominator 0? let's check,
x+2 = 0
x = -2
so, if "x" ever becomes -2, then you'd get
so, the domain, or values "x" can take on safely, are any real numbers EXCEPT -2.
Answer:
x is greater than or equal to 45
Step-by-step explanation:
