Answer:
(-2 , 5)
(-1 , 0)
(1 , -4)
(3 , 0)
(4 , -5)
Step-by-step explanation:
<u>First solve the equation:</u>
x² - 2x - 3
<em><u>Find two numbers with have a sum of -2 and a product of -3.</u></em>
-3 and 1
(x - 3)(x + 1)
Solve for x:
x - 3 = 0
x = 3
x + 1 = 0
x = -1
You know that the graph will cross the x-axis at -1 and 3.
(-1 , 0)
(3 , 0)
You know that the graph is positive.
<u>Complete the square to find the vertex</u>
x² - 2x - 3
(x - 1)² = x² - 2 + 1
x² - 2x - 3 = x² - 2 + 1 - 2 = (x - 1)² - 2
1 = 0
x = 1
Substitute into the original equation:
x² - 2x - 3 =
1² - (2 * 1) - 3 =
1 - 2 - 3 =
-4
(1 , -4)
<em><u>You can input any two numbers within -10 and 10. Such as -2 and 4.</u></em>
x² - 2x - 3 =
-2² - (2 * -2) - 3 =
4- -4- 3 =
5
(-2 , 5)
x² - 2x - 3 =
4² - (2 * 4) - 3 =
16 - 8 - 3 =
-5
(4 , -5)
(2x+4)+(4x+2)+(2x-1)=180
8x+5=180°
8x=180-5
8x=175
X=21.875
A= 2(21.875)+4=47.75
B=4(21.875)+2 =89.5
C=2(21.875)-1=42.75
Answer:
12 + 4x
Step-by-step explanation:
To solve this you would have to use the distributive property.
The problem would basically go like this.
4 (3+x)
(Distribute)
(4x3) (4 x X)
4x3 = 12, 4 x x = 4x
So the answer would be 12 +4x
Hope this helps :)
Answer:
Step-by-step explanation:
I see you're in college math, so we'll solve this with calculus, since it's the easiest way anyway.
The position equation is
That equation will give us the height of the rock at ANY TIME during its travels. I could find the height at 2 seconds by plugging in a 2 for t; I could find the height at 12 seconds by plugging in a 12 for t, etc.
The first derivative of position is velocity:
v(t) = -3.72t + 15 and you stated that the rock will be at its max height when the velocity is 0, so we plug in a 0 for v(t):
0 = -3.72t + 15 and solve for t:\
-15 = -3.72t so
t = 4.03 seconds. This is how long it takes to get to its max height. Knowing that, we can plug 4.03 seconds into the position equation to find the height at 4.03 seconds:
s(4.03) = -1.86(4.03)² + 15(4.03) so
s(4.03) = 30.2 meters.
Calculus is amazing. Much easier than most methods to solve problems like this.
