I believe the equation is y=350+150x with y being the total amount of money in his account and x being the number of months..Paul starts with $350, which is the y-intercept (starting value) of the equation then the slope is 150 because his total savings increases by $150 for every month he saves without making any withdrawals. I apologize if I'm wrong but I hope this helps.
Find the total cost of producing 5 widgets. Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $15.50 to produce 3 widgets, $23.50 to produce 7 widgets, and $56 to produce 12 widgets.
OK...so we have
a(7)^2 + b(7) + c = 23.50 → 49a + 7b + c = 23.50 (1)
a(3)^2 + b(3) + c = 15.50 → 9a + 3b + c = 15.50 subtracting the second equation from the first, we have
40a + 4b = 8 → 10a + b = 2 (2)
Also
a(12)^2 + b(12) + c = 56 → 144a + 12b + c = 56 and subtracting (1) from this gives us
95a + 5b = 32.50
And using(2) we have
95a + 5b = 32.50 (3)
10a + b = 2.00 multiplying the second equation by -5 and adding this to (3) ,we have
45a = 22.50 divide both sides by 45 and a = 1/2 and using (2) to find b, we have
10(1/2) + b = 2
5 + b = 2 b = -3
And we can use 9a + 3b + c = 15.50 to find "c"
9(1/2) + 3(-3) + c = 15.50
9/2 - 9 + c = 15.50
-4.5 + c = 15.50
c = 20
So our function is
c(x) = (1/2)x^2 - (3)x + 20
And the cost to produce 5 widgets is = $17.50
<span>A ball is thrown at a 30 degree angle above the horizontal with a speed of 10 ft/s. After 0.50s the horizontal component of it's velocity will be the same. In a projectile motion the horizontal component of the velocity is said to be constant. Therefore, it will be equal to the initial velocity.</span>
To find the future value the formula is
A=p e^rt
A future value?
P present value 8906.54
R interest rate 0.06
T time 9 years
E constant
A=8,906.54×e^(0.06×9)
A=15,283.68
To find the interest earned the formula is
I=A-p
I=15,283.68−8,906.54
I=6,377.14
=48
Multiply by 1
Add 1 1 11 11
to both sides of the equation
