9514 1404 393
Answer:
k = 2/15
Step-by-step explanation:
We can solve the given equation for k:
k = s/p . . . . . . divide the given equation by p on both sides
Using the first values from the table (15 pounds, 2 scoops), we have ...
k = 2/15
The value of k is 2/15.
<span>it depends how the interest is calculated, but there's not much of a difference
assuming its continuously compouned, you use this formula: A(t)=Pe^(rt), where A is the final amount, P is the initial investment, r is the interest, and t is the time in years
you want to find t such that A(t)=18,600 so 18,600=1000e^(.0675t)
you need to use logarithm to figure it out, take the natural log of both sides
the following properties will come into use:
ln(a*b)=ln(a)+ln(b)
ln(a^b)=bln(a)
ln(e)=1
taking the natural log
ln(18,600)=ln(1000e^(.0675t))
ln(18,600)=ln(1000)+ln(e^.0675t)
ln(18600)=ln(1000) + .0675t
now solve for t: t= (ln(18600)-ln(1000))/.0675
t=43.31</span>
Answer:
<h2>14mph</h2>
Step-by-step explanation:
Given the gas mileage for a certain vehicle modeled by the equation m=−0.05x²+3.5x−49 where x is the speed of the vehicle in mph. In order to determine the speed(s) at which the car gets 9 mpg, we will substitute the value of m = 9 into the modeled equation and calculate x as shown;
m = −0.05x²+3.5x−49
when m= 9
9 = −0.05x²+3.5x−49
−0.05x²+3.5x−49 = 9
0.05x²-3.5x+49 = -9
Multiplying through by 100
5x²+350x−4900 = 900
Dividing through by 5;
x²+70x−980 = 180
x²+70x−980 - 180 = 0
x²+70x−1160 = 0
Using the general formula to get x;
a = 1, b = 70, c = -1160
x = -70±√70²-4(1)(-1160)/2
x = -70±√4900+4640)/2
x = -70±(√4900+4640)/2
x = -70±√9540/2
x = -70±97.7/2
x = -70+97.7/2
x = 27.7/2
x = 13.85mph
x ≈ 14 mph
Hence, the speed(s) at which the car gets 9 mpg to the nearest mph is 14mph
