Let the required equation be y = mx + c; where y = 1, m = f'(0), x = 0
f(a) = sec(a)
f'(a) = sec(a)tan(a)
f'(0) = sec(0)tan(0) = 0
y = mx + c
1 = 0(0) + c
c = 1
Therefore, required equation is
y = 1.
Answer:
A. <u>4</u> trucks will make $80 because 4 × 20 = 80, so they need to wash <u>38</u> cars to make the other $380 because 38 × 10 = 380.
($80 + $380 = $460)
B. <u>15</u> trucks will make $300 because
15 × 20 = 300, so they need to wash <u>16</u> cars to make the other $160 because 16 × 10 = 160. ($300 + $160 = $460)
C. <u>21</u> trucks will make $420 because
21 × 20 = 420, so they need to wash <u>4</u> cars to make the other $40 because 4 × 10 = 40.
($420 + $40 = $460)
D. <u>27</u> trucks will make $540, so they won't have to wash any cars because they would have already exceeded their goal of $460.
Not sure what to do for letter E. Sorry. I hope the rest makes sense though :)
The simplest form of 204/72
=17/6
When you have this type of problem, you need to combine the like-terms and isolate the variable.
3x + 122 = 22x - 11
Add 11 to both sides to get rid of it
3x + 122 + 11 = 22x - 11 + 11 (-11 + 11=0)
3x + 133 = 22x
Then you would bring the 3x to the other side, so subtract 3x from both sides
3x + 133 = 22x
-3x -3x
133 = 22x - 3x
133 = 19x
Then divide both sides by 19 to isolate x
133/19 = 19x/19
133/19 = 7, so x = 7
Hope this helps!!
