6.2w = 18.6....divide both sides by 6.2
w = 18.6 / 6.2
w = 3 <==
Answer:
3/x + 2/(x+1) = 3/5x
[3 · 5(x + 1) + 2 · 5x] / 5x(x + 1) = 3(x + 1) / 5x(x + 1)
3 · 5(x + 1) + 2 · 5x = 3(x + 1)
15x + 15 + 10x = 3x + 3
15x + 10x - 3x = -15 + 3
22x = -12
x = -12/22 = -6/11
Answer:
2. a and b only.
Step-by-step explanation:
We can check all of the given conditions to see which is true and which false.
a. f(c)=0 for some c in (-2,2).
According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.
b. the graph of f(-x)+x crosses the x-axis on (-2,2)
Let's test this condition, we will substitute x for the given values on the interval so we get:
f(-(-2))+(-2)
f(2)-2
-1-1=-3 lower limit
f(-2)+2
1+2=3 higher limit
according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.
c. f(c)<1 for all c in (-2,2)
even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.
The final answer is then 2. a and b only.
It’s 2 and 3 bc 6a+12 and then 3(2a+4) and you need to distribute for that so 3 times 2a equals 6a and 3 times 4 equals 12 so you get 6a+12