Answer:
The result is the number of years, approximately, it'll take for your money to double. For example, if an investment scheme promises an 8% annual compounded rate of return, it will take approximately nine years (72 / 8 = 9) to double the invested money
Answer:
Our unknown, "a number," will be represented by "x." "The quotient of five times a number and 7" translates to 5x/7. "is no more than" is less than, which we will show using the symbol "<." 10 will remain 10. So, our inequality is 5x/7 < 10
Step-by-step explanation:
Each pound of the dog food costs $1.03 .
Each pound of the cat food costs $1.02 .
If you own a dog, then the dog food is the better deal.
If you own a cat, then the cat food is the better deal.
You would not want to feed your pet on food formulated for
a different species of animal, just to save a few pennies.
For that matter, I'll bet you could save a TON of money by
feeding your whole family dog food or cat food three times
a day !
So since the vertex falls onto the axis of symmetry, we can just solve for that to get the x-coordinate of both equations. The equation for the axis of symmetry is , with b = x coefficient and a = x^2 coefficient. Our equations can be solved as such:
y = 2x^2 − 4x + 12:
y = 4x^2 + 8x + 3:
In short, the vertex x-coordinate's of y = 2x^2 − 4x + 12 is 1 while the vertex's x-coordinate of y = 4x^2 + 8x + 3 is -1.
Answer:
year 7
Step-by-step explanation:
If we assume that investment A earns interest compounded annually, its value can be modeled by the equation ...
A = 50·(1+0.08)^(t-1) . . . . . where t is the year number
The second investment earns $3 per year, so its value can be modeled by the equation ...
B = 60 + 3(t -1) . . . . . . . . . where t is the year number
We are interested in finding the minimum value of t such that ...
A > B
50·1.08^(t-1) > 60 +3(t-1)
This is a mix of exponential and polynomial terms for which no solution method is available using the tools of Algebra. A graphing calculator shows the solution to be ...
t > 6.552
The value at the end of year 1 is found for t=1, so the values of interest are seen after 6.55 years, in year 7.