Hello from MrBillDoesMath!
Answer:
x < -2
Discussion:
7 < 3 - 2x => Add 2x to both sides
7 + 2x < 3 - (2x - 2x) => As 2x -2x = 0
7 + 2x < 3 => Subtract 7 from both sides
(7-7) + 2x < 3 - 7 => As 7-7 = 0 and 3-7 = -4
2x < -4 => Divide both sides by 2
x < -4./2 = -2
Thank you,
MrB
ANSWER
and
e have
EXPLANATION
Let us make y the subject and call it equation (2)
We put equation (2) in to equation (1)
Simplify to get,
Divide both sides by 31,
We put this value in to equation (2) to get,
We collect LCM to obtain,
Answer:
Step-by-step explanation:
9 - x ≤17
At some point you are going to have to turn the equation around. This would not normally be your first step, but this time it is better to start with it.
We won't do it directly. The best way to do it is to add x to both sides before you do anything else. This is not the usual way to solve these equations, but it's a good time to learn.
Inequality Rule: you must always solve for x. If it is -x then you are going to have to make an adjustment to get the x to be positive.
9 - x ≤ 17 Add x to both sides
9 - x+x ≤ 17 + x Combine
9 ≤ 17 + x Subtract 17 from both sides.
9 - 17 ≤ 17 - 17 + x
8 ≤ x
Notice that you have effectively changed the ≤ sign around, not because you have, but because the x reads differently now. It started out 9 - x ≤ 17 and when you finish solving it you get 8 is less than or equal to x. Entirely different.
A. √(0.8^2) + (0.6^2) = √1 = 1 => OK
<span>b.(-2/3,√ 5/3) = √(-2/3)^2 + 5/9) = √(4/9 +5/9) = √1 = 1 => OK
c.(√ 3/2, 1/3) = √(3/4 + 1/9) < 1 => it is inside the unit circle
d.(1,1)
= √(1 + 1) = √2 > 1 => NO. This point is beyond the limits of the unit circle.</span>
Answer:
m=-2
Step-by-step explanation:
As the product of the roots of a quadratic equation is c/a in ax^2+bx+c=0
here a=2, b=+8, c=-m^3
Given c/a=4
-m^3/2=4
-m^3=8
m^3= -8
m=-2.
