If you have a quadratic equation in the form ax^2+bx+c
Step 1) Determine the product of AC (the coefficients in a quadratic equation)
Step 2) Determine what factors of a⋅ca⋅c sum to bb
Step 3) "ungroup" the middle term to become the sum of the factors found in step 2
Step 4) group the pairs.
Just replace X with the given value.
X = -3
f(-3) = 5 - (-3)
f(-3) = 5 +3
f(-3) = 8
Answer:
-30/15 or -2
Step-by-step explanation:
2+8 = 10
3+2 = 5
14-2 = 12
4-1 = 3
10/5 - 12/3
(Find GCF)(GCF = 15)(5 times 3)
30/15 (10/5 times 3)
60/15 (12/3 times 5)
30/15 - 60/15 = -30/15 --> -2(simplified to 2)
Answer:
A) (1 s, 2.3 s)
B) (-4 m/s², 3.8 m/s²)
Step-by-step explanation:
The car's position which is the distance is given by the equation;
s(t) = t³ - 5t² + 7t
A) Velocity is the first derivative of the distance. Thus;
v(t) = ds/dt = 3t² - 10t + 7
At v = 0, we have;
3t² - 10t + 7 = 0
Using quadratic formula, we have;
t = 1 and t = 2.3
Thus, time at velocity of 0 is t = (1 s, 2.3 s)
B) acceleration is the derivative of the velocity. Thus;
a(t) = dV/dt = 6t - 10
At velocity of 0, we got t = 1 and t = 2.3
Thus;
a(1) = 6(1) - 10 = -4 m/s²
a(2.3) = 6(2.3) - 10 = 3.8 m/s
Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)
Answer:
The answers are:
A. Exactly one circle can be inscribed in a given triangle.
C. Many triangular shapes can be circumscribed about a given circle.
Step-by-step explanation:
I just took the quiz as well. Good luck!
