Reminder: y = mx + b, to isolate y, you would need to subtract mx value, and what you do to one side, you do to all. If there’s a coefficient in front of your y value, you then divide the equation by that coefficient. Once y is isolated, m and b values are easier to find.
Answer:
1) y = -2x + 1 // m = -2, b = 1
2) y = 5/8x - 1 // m = 5/8, b = -1
3) y = 7/8x - 3/4 // m = 7/8, b = -3/4
4) y = 3/4x + 4 // m = 3/4, b = 4
Answer:
Step-by-step explanation:
The letters are easy enough.
y^3 determines that 3 ys are needed. y^2 is taken in by the cubed.
x^4 determines that 4 xs are needed. x^3 is part of x^4 and you don't need any more xs
12 and 42 are the parts that will cause the problem. Factor them both into prime factors
12: 2 * 2 * 3
42: 2 * 3 * 7
You need two 2s.
You need one 3
You need one 7
LCD = 2 *2*3 * 7 = 84
To solve this, you must rearrange the variables in the given equation. Since we are looking for h, we want the equation to be h = something. You must divide both sides by 1/3 lw. When dividing like this, you must use the KCF (Keep, Change, Flip) method. Keep the first part (v), change the sign from division to multiplication, and flip the fraction. This means that the fraction becomes 3/1 lw. When that is multiplied by v, it makes the equation h = 3v/lw. That is your final answer -> h = 3v/lw.
Im pretty sure it’s the last one.
5^-3 = 1/125
-5^-3 = -1/125
(-5^-3)^-1 = -125
(-5^-3)^0 = 1
