Here is a quadratic equation x^2 + 5x = 14
x^2 + 5x - 14 = 0 - Subtract 14 from both sides
(x + 7)(x - 2) = 0 - Factor
x + 7 = 0 x - 2 = 0 - Set each factor equal to 0
x = -7 x = 2 - solve each piece
If the constant is easy to find factors of then use the factoring method
If the constant is difficult and you have a graphing calculator, then graph
If no graphing calculator then use the quadratic formula
When the ball hits the ground, the height is 0.
0 = 100 - 16t²
0 = (10 - 4t)(10 + 4t)
0 = 10 - 4t or 0 = 10 + 4t
4t = 10 or -4t = 10
t =
or t = 
Time cannot be negative (unless you have a time machine) so disregard 
Answer: t =
= 2.5 seconds
Answer:
We need to use
Step-by-step explanation:
We have to use the equation for passing through the points known as
y2-y1/x2-x1
-1-3/0-5= -4/-5
the answer is 4/5
decimal: 0.8
The equation of line is y = 92x – 182631. The slope is 92 and the y-intercept is the negative 182631. And the health expenditure in 2010 will be 2289.
<h3>What is the equation of a line passing through two points?</h3>
Suppose the given points are (x₁, y₁) and (x₂, y₂), then the equation of the straight line joining both two points is given by
![\rm (y - y_1) = \left [ \dfrac{y_2 - y_1}{x_2 - x_1} \right ] (x -x_1)](https://tex.z-dn.net/?f=%5Crm%20%28y%20-%20y_1%29%20%3D%20%5Cleft%20%5B%20%5Cdfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D%20%5Cright%20%5D%20%28x%20-x_1%29)
In MODE, (second line), fix the number of decimal places to be 2.
Then the slope and the y-intercept values of the Lin Reg line will round to the nearest hundredth.
(x₁, y₁) → (1997, 1093)
(x₂, y₂) → (2002, 1553)
Then we have
y – 1093 = [(1553 – 1093) / (2002 – 1997)] (x – 1997)
y – 1093 = [460 / 5] (x – 1997)
y – 1093 = 92 (x – 1997)
y = 92x – 182631
If x = 2010, then the value of y will be
y = 92 × 2010 – 182631
y = 184920 – 192631
y = 2289
Learn more about straight-line equations here:
brainly.com/question/380976
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The correct answer should be A because you have to have a common denominator in order to solve so you take 5/6x and 4/6x and solve.