PLS HELP ITS EMERGENCY
1 answer:
You might be interested in
Answer:
There are two pairs of solutions: (2,7) and (-1,4)
Step-by-step explanation:
We will use substitution.
y = x^2 + 3
y = x +5
Since the second equation is equal to y, replace y in the first equation with the second equation.
y = x^2 + 3
x + 5 = x^2 + 3
Rearrange so that one side is equal to 0.
5 - 3 = x^2 - x
2 = x^2 - x
0 = x^2 - x - 2
You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).
a = 1, b = -1, c = -2
Zeros = and
Zeros = 2 and -1
Now that you have your x values, plug them into the equations to find their corresponding y values.
y = x^2 + 3
y = (2)^2 + 3
y = 7
Pair #1: (2,7)
y = x^2 + 3
y = (-1)^2 + 3
y = 4
Pair #2: (-1,4)
Therefore, there are two pairs of solutions: (2,7) and (-1,4).
Hello there! We are given the following equation and want to simplify it:
First, let's use the Distributive Property:
For this step, we are going to let:
After applying this version of the Distributive Property, we are given the equation:
We're going to use the distributive party again, but a slightly different version that's almost the same as first:
Using this on both terms of the equation changes the equation to:
Now, we just remove the parenthesis and combine like terms
This should be your answer. Let me know if you need any clarifications, thanks!
~ Padoru
Answer:
s = - 1 ±
Step-by-step explanation:
Given
s² + 2s - 6 = 0 ( add 6 to both sides )
s² + 2s = 6
To complete the square
add ( half the coefficient of the s- term )² to both sides
s² + 2(1)s + 1 = 6 + 1
(s + 1)² = 7 ( take the square root of both sides )
s + 1 = ± ( subtract 1 from both sides )
s = - 1 ±
Thus
s = - 1 - , s = - 1 +