Answer:
C. d = 24g
Step-by-step explanation:
The problem boils down to determining the ratio between d and g. That is, for some equation ...
d = k·g
we want to determine the value of k. Solving the equation for that value, we find ...
k = d/g
So, we need only to read a point from the graph with sufficient accuracy to determine a good estimate for k.
(gallons, miles) = (g, d) = (5, 120) is a suitable point
Then ...
k = d/g = 120/5 = 24
The equation is d = 24g.
Answer:
y=7600(5^(t/22))
Step-by-step explanation:
This is going to be an exponential function as it grows rapidly.
This type of question can be solved using the formula y=a(r^x), where a is the inital amount, r the factor by which the amount increases and x is the unit of time after which the amount increases.
x=t/22
a=7600
r=5
∴y=7600(5^(t/22))
I would half 198, get
99 and add a 0, this is how you multiply by 5 in your head
198*5 = 990
990 + 200 = 1190 - 2
= 1188
Alternatively we can round 198 to 200: then multiply by 6
200*6 = 1200
Then We subtract the 6 "2's" that we added
1200 - 12 = 1188
The expression for this is:
(6*200) - (6 * 2)
Answer: SSS
Proof:
In ΔMLQ and ΔNPQ,
MQ = NQ (given) S
Since Q is the midpoint of LP, by definition, LQ = QP (S)
LM = PN (given) S
∴ ΔMLQ ≡ ΔNPQ (SSS)
