Consider the charge for parking one car for t hours.
If t is more than 1, then the function is y=3+2(t-1), because 3 $ are payed for the first hour, then for t-1 of the left hours, we pay 2 $.
If t is one, then the rule y=3+2(t-1) still calculates the charge of 3 $, because substituting t with one in the formula yields 3.
75% is 75/100 or 0.75.
For whatever number of hours t, the charge for the first car is 3+2(t-1) $, and whatever that expression is, the price for the second car and third car will be
0.75 times 3+2(t-1). Thus, the charge for the 3 cars is given by:
3+2(t-1)+0.75[3+2(t-1)]+0.75[3+2(t-1)]=3+2(t-1)+<span>0.75 × 2[3 + 2(t − 1)].
Thus, the function which total parking charge of parking 3 cars for t hours is:
</span><span>f(t) = (3 + 2(t − 1)) + 0.75 × 2(3 + 2(t − 1))
Answer: C</span>
i. 171
ii. 162
iii. 297
Solution,
n(U)= 630
n(I)= 333
n(T)= 168
i. Let n(I intersection T ) be X
<h3>ii.
n(only I)= n(I) - n(I intersection T)</h3><h3>
= 333 - 171</h3><h3>
= 162</h3>
<h3>
iii. n ( only T)= n( T) - n( I intersection T)</h3><h3>
= 468 - 171</h3><h3>
= 297</h3>
<h3>
Venn- diagram is shown in the attached picture.</h3>
Hope this helps...
Good luck on your assignment...
Answer:
x = 21
Step-by-step explanation:
<u>Step 1: Make an expression</u>
Since they are supplements, that means that they add up to 180 degrees.
3x + 6 + 5x + 6 = 180
<u>Step 2: Combine like terms</u>
3x + 6 + 5x + 6 = 180
(3x + 5x) + (6 + 6) = 180
8x + 12 = 180
<u>Step 3: Subtract 12 from both sides</u>
8x + 12 - 12 = 180 - 12
8x = 168
<u>Step 4: Divide both sides by 8</u>
8x / 8 = 168 / 8
x = 21
Answer: x = 21
Answer:
64
Step-by-step explanation:
its just how it is
F(g(-2))=4(-2)^6+4(-2)^3+1
f(g(-2))=4(64)+4(-8)+1
f(g(-2))=256-32+1
f(g(-2))=224+1
f(g(-2))=225
