Answer:
Owen can afford to keep and drive the car for 4 days.
Step-by-step explanation:
The total owing on the car rental is R(d) = ($77.25/day)d + ($0.12/mile)m ≤ $330. Substitute 1 for d (that is, the rental is for 1 day) and 175 for m:
R(d) = ($77.25/day)(d days) + ($0.12/mile)(175 miles) ≤ $330
= $77.25d + $21 ≤ $330
This simplifies to $77.25d + $21 ≤ $330, or
$77.25d ≤ $309
Solving this by dividing both sides of the above equation by $77.25, we get
d = ($309)/($77.25) = 4
Answer:
The number of tickets sold to the public was 375
Step-by-step explanation:
<u><em>The complete question is</em></u>
At a charity basketball game, 450 tickets were sold to students at a school. The remaining x tickets were sold to the public. The students paid $15 and public $25. When all the tickets were sold, $10,500 was collected. How many tickets were sold to the public?
Let
x ----> number of tickets sold to the students
y ----> number of tickets sold to the public
we know that
----> equation A
----> equation B
Solve the system by graphing
The solution is the intersection point both graphs
using a graphing tool
The solution is the point (75,375)
see the attached figure
therefore
The number of tickets sold to the public was 375
I would say to check the 1st and 3rd
Answer:
a)z1 +z2 =z2 + z1 ...proved.
b) z1 + ( z2+ z3 )=(z1+z2)+z3 ... proved.
Step-by-step explanation:
It is given that there are three vectors z1 = a1 + ib1, z2 = a2 + ib2 and z3 = a3 + ib3. Now, we have to prove (a) z1 + z2 = z2 + z1 and (b) z1 + (z2 +z3) = (z1 + z2) + z3.
(a) z1 + z2 = (a1 +ib1) + (a2+ ib2) = (a1 +a2) + i(b1 +b2) {Adding the real and imaginary parts separately}
Again, z2 + z1 =(a2 +ib2) + (a1 +ib1) = (a2 +a1) + i(b2 +b1) {Adding the real and imaginary parts separately}
Hence, z1 +z2 =z2 + z1 {Since, (a1 +a2) = (a2 +a1) and (b1 +b2) = (b2 +b1)}
(b) z1 + ( z2+ z3 ) = [a1 + ib1] + [(a2 + a3 ) + i(b2 + b3 )] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}
Again, (z1+z2)+z3 = [(a1+a2) +i(b1+b2)]+[a3+ib3] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}
Hence, z1 + ( z2+ z3 )=(z1+z2)+z3 proved.
