y = x2 - 3x - 20 is a formula linking two variables, x and y. It does not - indeed, it cannot - have a solution.
The equation x2 - 3x - 20 = 0 has solutions at x = -3.217 and x = 6.217 approx.
Solve, using the Rule of 72 rate = 4%, years = 18, fv=$8,000. Solve for PV. Formula: PV = $1/(1+r) t PV = $8000/(1+.04) 18 PV = $8000/2.0258 3949.03 = $8000/2.20258
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
(2,-2)
Since the second equation is already solved for "y", you can replace "y" by "9" in the other equation. Then solve the new equation for "x".
You cannot solve one linear equation in two variables. You need two equations that are independent.
x^(2) - 10x + 16 = 0 Factors of '16', which are, 1,2,4,8,16. Select two numbers from this list that add/subtract to '10' . They are '2' and '8'. Setting up brackets ( x 2)(x 8) Since '16' is positive(+), then both signs must be the same. Since '10x' is negative (-) , then both signs are negative. Hence (x - 2)(x - 8 ) = 0 When x - 2 = 0 x = 2 & when x - 8 = 0 x = 8 So the two answers are ,2, & ,8, .
8840-026
Solve, using the Rule of 72 rate = 4%, years = 18, fv=$8,000. Solve for PV. Formula: PV = $1/(1+r) t PV = $8000/(1+.04) 18 PV = $8000/2.0258 3949.03 = $8000/2.20258
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
the answer
(2,-2)
You can solve lineaar quadratic systems by either the elimination or the substitution methods. You can also solve them using the comparison method. Which method works best depends on which method the person solving them is comfortable with.
Since the second equation is already solved for "y", you can replace "y" by "9" in the other equation. Then solve the new equation for "x".
a=5: c=4
a2+30a+56=0 , solve for a Using the quadratic formula, you will find that: a=-2 , a=-28
You cannot solve one linear equation in two variables. You need two equations that are independent.
"tan times 5A2" makes no sense.