1. Regroup terms
x^3 - 6 * 10x^2x-8/ x-3
2. Use product rule
x^3 - 6 * 10x^ 2+1 - 8/ x-3
3. Simplify
x^3 - 6 * 10 x^3 - 8/ x-3
4. Simplify further
x^3 - 60x^3 - 8/ x-3
5. Simplify the last time
-59x^3-8/x-3
The solutions fo the inequality are all the points (x, y) that meet these 3 conditions.
- x ≠ 0
- y ≠ 0
- Sign(x) =sign(y)
<h3>
Which points are solutions of the inequality?</h3>
We want to find points of the form (x, y) that are solutions of the inequality:
x*y > 0
Clearly x and y must be different than zero.
So, for example if x = -1, y can be any negative number, for example y= -3
x*y > 0
(-1)*(-3) > 0
3 > 0
This is true.
Now if x = 1, y must be positive. LEt's take y = 103, then:
x*y > 0
1*103 > 0
103 > 0
Then we have 3 conditions:
- x ≠ 0
- y ≠ 0
- Sign(x) =sign(y)
The solutions fo the inequality are all the points (x, y) that meet these 3 conditions.
If you want to learn more about inequalities:
brainly.com/question/25275758
#SPJ1
Whatever number is the exponent if negative just move it to the right how much the number is. if it 5 with an exponent of -3 then you just move the decimal point 3 times to the right. The anwer to that question would be .005
Answer:
385 golf balls
Step-by-step explanation:
margin of error = (z*)(s) / sqrt n
where z* = 1.96 with 5%/ 2 = 0.025 area in each tail
margin of error = (z*)(s) / sqrt n
1.2 yards = (1.96)(12 yards) / sqrt n
solve for n
n = 384.16
385 golf balls (always round up)
Answer:
The correct option is;
B
Step-by-step explanation:
The given system of inequalities are;
5·x - 4·y > 4...(1)
x + y < 2...(2)
Representing both inequalities as a function of "y", gives;
For, 5·x - 4·y > 4...(1), we have;
-4·y > 4 - 5·x
y < 4/(-4) - 5·x/(-4)
∴ y < 5·x/4 - 1
For x + y < 2...(2), we have;
y < 2 - x
Therefore, y is less than the values given by the equation of the straight line equalities, and the feasible region is given by the common region under both dashed lines representing both inequalities as shown in the attached diagram created using Microsoft Excel
The correct option is therefore, B.
