8/13 = 0.6153
y = cos-1(0.6153)
y = 52.0
I'm not totally sure if this is what you mean or if this is.
y = cos(8/13) = 0.9999 which to the nearest tenth is 1.0
1000 is our constant because the truck was only used one day. We also know they charge a certain amount per hour, but we do not know how much. In this case they charged that amount 9 times because the truck was used for 9 hours. The total of the constant plus the 9hour fee is 2700 dollars:
1000(1) + 9x = 2700
The reason we multiply 1000 by 1, is because we only used the truck for one day.
Solve for x
9x = 2700 - 1000 = 1700
X = 1700/9 = 188.88
The hourly fee is $188.88
The question is asking for an equation for hours of use though, so the answer is
1000(1) + 9(188.88) = C
C represents cost or charge.
Answer:
a) 2,650%
b) 27,400%
c) 164,900%
Step-by-step explanation:
We want to measure the percentage we would have to increase the typical value to obtain the values given in a), b) and c).
But 0.2 increased in x% equals
0.2 + 0.2(x/100) =0.2(1+x/100)
So, if we want to increase 0.2 in x%, we must multiply it by (1+x/100)
a)
We need to find the value of x such that
0.2(1+x/100) = 5.5 ⇒ (1+x/100)=5.5/0.2 ⇒ 1+x/100=27.5
⇒ x/100=26.5 ⇒ x=2,650%
b)
0.2(1+x/100) = 55 ⇒ (1+x/100)=55/0.2 ⇒ 1+x/100=275
⇒ x/100=274 ⇒ x=27,400%
c)
5.5 min = 5.5*60 s = 330
0.2(1+x/100) = 330 ⇒ (1+x/100)=330/0.2 ⇒ 1+x/100=1,650
⇒ x/100=1,649 ⇒ x=164,900%
