Total investment=$2500
Let say David earned 5% profit and Harry earned 3% profit.
If <u>David</u> invested y dollar in mutual fund then <u>Harry</u> would invest,
Harry would invest,$(2500-y)
Total profit would equal to sum of 5% of money <u>David</u> invested and 3% of money <u>Harry</u> invested,
So,
5% × y +3% ×(2500-y)=<u>total profit</u>
0.05×y+0.03×(2500-y)=$111
0.05y+0.03×2500-0.03y=111
0.02y+75=111
0.02y=111-75
y=36/0.02
y=$1800
which is the amount <u>David</u> invested.
The amount invested by <u>Harry</u> would be,
=$(2500-1800)
=$700
1)
x^2 + 4x - 5
y = --------------------
3x^2 - 12
Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite
=> 3x^2 - 12 = 0 => 3x^2 = 12
=> x^2 = 12 / 3 = 4
=> x = +/-2
Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) = ∞
Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞
Limit of y when x -> - 2(+) = +∞
Limit of y when x -> - 2(-) = -
=> x = 2 and x = - 2 are a vertical asymptotes
Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite
Limi of y when x -> - ∞ = 1/3
Lim of y when x -> +∞ = 1/3
=> Horizontal asymptote y = 1/3
x - intercept => y = 0
=>
x^2 + 4x - 5
0 = ------------------ => x ^2 + 4x - 5 = 0
3x^2 - 12
Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1
=> x-intercepts x = - 5 and x = 1
Domain: all the real values except x = 2 and x = - 2
2) y = - 2 / (x - 4) - 1
using the same criteria you get:
Vertical asymptote: x = 4
Horizontal asymptote: y = - 1
Domain:all the real values except x = 4
Range: all the real values except y = - 1
3)
x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)
factor x^2 - 3x - 10 => (x - 5)(x + 2)
Multiply both sides by (x - 5) (x + 2)
=> x(x - 5) + 7( x + 2) = 14
=> x^2 - 5x + 7x + 14 = 14
=> x^2 + 2x = 0
=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0
Answer: x = 0
Answer:
see the explanation
Step-by-step explanation:
<u><em>The correct question is</em></u>
Use the discriminant to determine how many solutions are possible for the following equation (show work).
5x^2-3x+4=0
we know that
The discriminant for a quadratic equation of the form
is equal to
If D=0 then the equation has only one real solution
If D>0 then the equation has two real solutions
If D<0 then the equation has no real solutions (two complex solutions)
in this problem we have
so
substitute
so
The equation has no real solutions, The equation has two complex solutions
therefore
I know there are___No____
real solutions to the equation in problem 4 because ___the discriminant is negative___
Answer:
- no real solutions
- 2 complex solutions
Step-by-step explanation:
The equation can be rearranged to vertex form:
x^2 -4x = -5 . . . . . . . . . subtract 4x
x^2 -4x +4 = -5 +4 . . . . add 4
(x -2)^2 = -1 . . . . . . . . . show the left side as a square
x -2 = ±√-1 = ±i . . . . . . take the square root; the right side is imaginary
x = 2 ± i . . . . . . . . . . . . . add 2. These are the complex solutions.
_____
<em>Comment on the question</em>
Every 2nd degree polynomial equation has two solutions. They may be real, complex, or (real and) identical. That is, there may be 0, 1, or 2 real solutions. This equation has 0 real solutions, because they are both complex.
