Too bad you haven't shared illustrations of the possible solutions.
Starting with 4|x+3|>8, div. both sides by 4: <span>|x+3|>2
Case 1: x+3 is already positive: then x+3>2, or x > -1
Case 2: x+3 is negative: Then -(x+3)>2, or -x - 3 > 2, or -5>x or x<-5
Draw a number line representing x values. Place empty circles at x = -5 and x = -1. Draw a vector from the circle at x = -1, to the right of x = -1. Draw a vector from the circle at x = - 5, to the left of x = -5. Note that x values between x = -5 and x = -1 are not solutions.</span>
Answer:
5:2
Step-by-step explanation:
Start with what the question is asking for, which is the ratio of math problems to minutes.
With ratios "to" also means ":", so you can change it to math problems : minutes.
Then, substitute the numbers for the labels. Math problems = 10 and minutes = 4, so now you have 10:4
Lastly, to simplify the ratio, find the greatest common factor, in this case 2, and divide both sides, which gives you your answer 5:2
The answer is A : y=-1/5x+1 ..... I checked on desmos
Answer:
-1-1.6n
Step-by-step explanation:
5+3.2n-6-4.8n
add like terms
-1-1.6n
Answer:
50 units
Step-by-step explanation:
Find the number of units x that produces the minimum average cost per unit C in the given equation.
C = 0.001x³ + 5x + 250
unit cost f(x) = C/x
= 0.001x³/x + 5x/x+ 250/x
f(x) = 0.001x² + 5 + 250/x
f'(x) = 0.002x - 250/x²
We equate the first derivative to zero
0.002x - 250/x² = 0
0.002x = 250/x²
Cross Multiply
0.002x × x² = 250
0.002x³ = 250
x³ = 250/0.002
x³ = 125000
x = 3√(125000)
x = 50 units
Therefore, the number of units x that produces the minimum average cost per unit C is 50 units.
