Answer:
v=-5 and v=3
Step-by-step explanation:
We are given that
We have to find two solutions of quadratic equation.
Using addition property of equality
(By using factorization method)
Substitute each factor equal to 0
and 
and 
Hence, two solutions of quadratic equation are
v=-5 and v=3
D = T2 - T1
d = 9 - 14
d = - 5
T40 = 14 + 39(-5)
= 14 - 195
= - 181
B i hope this helped and I hope you pass
Given:
The vertices of a quadrilateral ABCD are A(0, 4), B(4, 1), C(1, -3), and D(-3, 0).
To find:
The perimeter of quadrilateral ABCD.
Solution:
Distance formula:
Using the distance formula, we get
Similarly,
And,
Now, the perimeter of the quadrilateral ABCD is:
Therefore, the perimeter of the quadrilateral ABCD is 20 units.
Original equation: x^2 + 3x - 4 = 0
How I solve by factoring is to take term a and c (as in if you were using the quadratic equation), then multiply them together. We want to find two numbers that multiply to equal the product of a x c, but add up to term b.
Term A: x^2 (1)
Term B: -4
1 x -4 = -4
Factors that have a product of -4 and add up to 3: 4, -1
Factored equation: (x + 4)(x - 1) = 0
Set each factored section equal to 0 and solve for x.
x + 4 = 0
x = -4
x - 1 = 0
x = 1
The correct answer is D. x = 1, -4
Hope this helps!! :)
