Answer:
if it is easy i can do it
hope it helps
<h3>
Answer: g = -2</h3>
======================================================
Explanation:
We will use the slope formula
m = (y2-y1)/(x2-x1)
Plugging in the given items leads to
-5 = (g-8)/(6-4)
Now solve for g
-5 = (g-8)/(6-4)
-5 = (g-8)/2
-5*2 = g-8
-10 = g-8
-10+8 = g
-2 = g
g = -2
We have (6,g) turn into (6,-2)
The two points (4,8) and (6,-2) are on a line of slope -5.
-------------------
As a way to check, we'll find the slope of the line through (4,8) and (6,-2)
m = (y2-y1)/(x2-x1)
m = (-2-8)/(6-4)
m = -10/2
m = -5
The answer is confirmed.
Y=-2x^2-3x+4
dy/dx=-4x-3, d2y/dx2=-4
When the velocity, dy/dx=0, then y(x) is at an absolute maximum because the acceleration is a constant negative, -4.
dy/dx=0 only when -4x-3=0, -4x=3, x=-3/4
y(-3/4)=5.125
So the maximum value of y is 5 1/8
The answer is: x = 7 - √53 or x = 7 + √53
The general quadratic equation is: ax² + bx + c =
0.
But, by completing the square we turn it into: a(x + d)² + e = 0, where:<span>
d = b/2a
e = c - b²/4a
Our quadratic equation is x² - 14x -4 = 0, which is
after rearrangement:
So, a = 1, b = -14, c = -4
Let's first calculate d and e:
d = b/2a = -14/2*1 = -14/2 = -7
e = c - b²/4a = -4 - (-14)</span>²/4*1 = -4 - 196/4 = -4 - 49 = -53<span>
By completing the square we have:
a(x + d)² + e = 0
1(x + (-7))</span>² + (-53) = 0
(x - 7)² - 53 = 0
(x - 7)² = 53
x - 7 = +/-√53
x = 7 +/- √53
Therefore, the solutions are:
x = 7 - √53
or
x = 7 + √53
