Answer:
y = -2x^2(x - 3)
Step-by-step explanation:
<em><u>Preliminary Remark</u></em>
If a cubic is tangent to the x axis at 0,0
Then the equation must be related to y = a*x^2(x - h)
<em><u>(3,0)</u></em>
If the cubic goes through the point (3,0), then the equation will become
0 = a*3^2(3 - h)
0 = 9a (3 - h)
0 = 27a - 9ah
from which h = 3
<em><u>From the second point, we get</u></em>
4 = ax^2(x - 3)
4 = a(1)^2(1 - 3)
4 = a(-2)
a = 4 / - 2
a = -2
<em><u>Answer</u></em>
y = -2x^2(x - 3)
Answer:
16 pieces of candy per bag
Step-by-step explanation:
Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)
Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)
Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer: {r = -17/26
{s = 23/13 {t = 21/26
Answer:
The solutions are x = 1.24 and x = -3.24
Step-by-step explanation:
Hi there!
First, let´s write the equation:
log[(x² + 2x -3)⁴] = 0
Apply the logarithm property: log(xᵃ) = a log(x)
4 log[(x² + 2x -3)⁴] = 0
Divide by 4 both sides
log(x² + 2x -3) = 0
if log(x² + 2x -3) = 0, then x² + 2x -3 = 1 because only log 1 = 0
x² + 2x -3 = 1
Subtract 1 at both sides of the equation
x² + 2x -4 = 0
Using the quadratic formula let´s solve this quadratic equation:
a = 1
b = 2
c = -4
x = [-b± √(b² - 4ac)]/2a
x = [-2 + √(4 - 4(-4)·1)]/2 = 1.24
and
x = [-2 - √(4 - 4(-4)·1)]/2 = -3.24
The solutions are x = 1.24 and x = -3.24
Have a nice day!
