All real solutions
Everything cancels out once you multiply the top equation by two.
0 = 0 which means all real answers.
I hoping you meant question 3. Good luck!
Answer:
Yes they can all be written in y = mx + b. You just have to move the terms around.
Step-by-step explanation:
y = 2x -3, this is already in slope-intercept form
Now, y - 2 = x + 2: We can add 2 on both sides to cancel out the one on the left side:
y - 2 = x + 2
y - 2 + 2 = x + + 2
y = x + 4 <-- This is in y = mx + b form
Now the last one, 3x = 9 + 3y
We can first divide all terms by 3,
3x = 9 + 3y
/3 /3 /3
x = 3 + y: Then we can subtract 3 from both sides:
x - 3 = 3 + y - 3
x - 3 = y
These are all linear equations because none of the x's have bigger powers than 1. x^2 is a quadratic equation and x^3 is cubic equation.
Answer:
The solution of the equations are -6 and 1
Step-by-step explanation:
* <em>Lets explain how to solve the problem</em>
- We want to find the solution of the equation (x + 2) (x + 3) = 12
- <em>At first lets use the Foil method to multiply the two brackets</em>
(x + 2) (x + 3) = (x)(x) + (x)(3) + (2)(x) + (2)(3)
(x + 2) (x + 3) = x² + 3x + 2x + 6 ⇒ add the like term
(x + 2) (x + 3) = x² + 5x + 6
∵ (x + 2) (x + 3) = 12
∴ x² + 5x + 6 = 12
- Subtract 12 from both sides
∴ x² + 5x - 6 = 0
- <em>Factorize the left hand side</em>
∵ x² = (x)(x)
∵ -6 = 6 × -1
∵ 6x + -1x = 5x
∴ (x + 6)(x - 1) = 0
- <em>Lets use the zero product property </em>
∵ (x + 6)(x - 1) = 0
∴ x + 6 = 0 ⇒ <em>OR</em> ⇒ x - 1 = 0
∵ x + 6 = 0
- Subtract 6 from both sides
∴ x = -6
∵ x - 1 = 0
- Add 1 to both sides
∴ x = 1
∴ The solution of the equations are -6 and 1
Answer:
2.7 kb
Step-by-step explanation:
Total kb = 7.48
Total apps = 2
one app = 4.78
another app = x
> 4.78 + x = 7.48
x = 7.48 – 4.78
= 2.7 kb