50t + .1t^2 = 1100
Solve. show steps please
I would multiply by 10 and re-arrange into a standard quadratic equation.
t^2 + 500t - 11000 = 0
now use the quadratic formula to find the two real roots.
Im in 9th grade algebra. and we are learning adding and subtracting real numbers. im stumped on a couple of problems. I get confused when i see the absolute value lines. Does that change the munber? Do i just go on with the problem as if they werent there?
And i have NO idea how to do 2/5 - 7/8?!?
To solve the equation 50t + 0.1t^2 = 1100, you can follow these steps:
Step 1: Arrange the equation in standard form.
0.1t^2 + 50t - 1100 = 0
Step 2: Multiply the entire equation by 10 to eliminate the decimal.
t^2 + 500t - 11000 = 0
Step 3: Use the quadratic formula to solve for t. The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 500, and c = -11000.
t = (-500 ± √((500)^2 - 4(1)(-11000))) / (2 * 1)
t = (-500 ± √(250000 + 44000)) / 2
t = (-500 ± √(294000)) / 2
t = (-500 ± 540) / 2
Step 4: Calculate the two solutions.
t1 = (-500 + 540) / 2 = 40 / 2 = 20
t2 = (-500 - 540) / 2 = -1040 / 2 = -520
So, the solutions to the equation are t = 20 and t = -520.