Assuming that you’re asking for a fraction of 0.31 where 31 repeats?
31/99
Answer:
#5 39 = x
Sides:
x - 15
2x
2x
Total of sides = 180
180 = (x-15) + 2x + 2x
180 = x - 15 + 4x
180 = -15 + 5x (add 15 to each side to remove -15)
195 = 5x (divide by 5 on each side to get x)
39 = x
#7 24 = x
Sides:
2x - 4
3x + 5
2x + 11
Total of sides = 180
180 = (2x - 4) + (3x + 5) + (2x + 11)
180 = 2x - 4 + 3x + 5 + 2x + 11
180 = 7x + 12 (subtract 12 to each side to remove 12)
168 = 7x (divide by 7 on each side to get x)
24 = x
#8 37 = x
Sides:
x
x + 4
3x - 9
Total of sides = 180
180 = (x) + (x + 4) + (3x - 9)
180 = x + x + 4 + 3x - 9
180 = 5x - 5 (add 5 to each side to remove -5)
185 = 5x (divide by 5 on each side to get x)
37 = x
Step-by-step explanation:
12x12= 144, 144x24 = 3,456. 3,456 is the answer
Answer:
The proof is detailed below.
Step-by-step explanation:
We will first prove that if H(x) is a differentiable function in [a,b] such that H'(x)=0 for all x∈[a, b] then H is constant. For this, take, x,y∈[a, b] with x<y. By the Mean Value Theorem, there exists some c∈(x,y) such that H(y)-H(x)=H'(c)(x-y). But H'(c)=0, thus H(y)-H(x)=0, that is, H(x)=H(y). Then H is a constant function, as it takes the same value in any two different points x,y.
Now for this exercise, consider H(x)=F(x)-G(x). Using differentiation rules, we have that H'(x)=(F-G)(x)'=F'(x)-G'(x)=0. Applying the previous result, F-G is a constant function, that is, there exists some constant C such that (F-G)(x)=C.