Solve. -1/8c=2
a. 4
b. –16
c. –4
d. 16
Answers
This requires that c ≠ 0 , meaning c≠0
Multiply by x to eliminate var in denominator
Common denominator is 8
The answer is: b. -16
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Task 4—Solving Rational EquationsUsing the equation below as a model, fill in numbers in the place of a and b to create a rational equation that has an extraneous solution. x + a (over(fraction)) ax = b (over(fraction)) x Part 1. Show all work to solve for x in the equation and check the solution. Part 2. Explain how to identify the extraneous solution and what it means.
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Answer: $200.
Any firm (regardless of market structure) will maximize profit by producing and selling the quantity at which marginal revenue is equal to marginal cost.In the special case of a perfectly competitive firm,marginal revenue is equal to price.
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Answer:x=2/3 =0.6667
Step-by-step explanation:
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Answer:
Unit price= $4.30 create equation x= 4.50y
where x= total cost and y= $4.30
solve by multiplying $4.30x 4.50
x=$19.35
final answer: the total cost of the material was $19.35
Step-by-step explanation:
Answers
Answer:
A box-and-whisker plot shows the scores on a math exam for two classes.
Class A Class B
1) Minimum value 65 56
2) Lower quartile 66 79
3) Median 81 91
4) Upper quartile 89 94
5) Maximum value 95 100
First quartile also known as Lower quartile is the number below which lies the 25 percent of the bottom data.
Second quartile or Median divides the range in the middle and has 50 percent of the data below it.
Third quartile also known as upper quartile has 75 percent of the data below it and the top 25 percent of the data above it.
Interquartile Range = Upper quartile - Lower quartile,
For class A
Interquartile range = 89 -66 = 23
For class B
Interquartile range = 94 -79 = 15
The interquartile ranges tell you about the two classes that Class B has more consistent scores
Class A Class B
1) minimum value 65 56
2) 1st quartile 66 79
3) median 81 91
4) 3rd quartile 89 94
5) maximum value 95 100
Interquartile range = 3rd Quartile - 1st Quartile
Class A: 89 - 66 = 23
Class B: 94 - 79 = 15
IQR indicates the spread of the of the scores over the median. Class A has a spread that his more than 50% of the scores of Class B.
Answers
Answer: B
Step-by-step explanation:
Looking at the inequalities, both are ≤ and ≥. That means the x is not only greater or less than, but they are also equal to the number. With this fact, we can eliminate D since D has open circles.
Orr problem has 2 separate inequalities. Both are doing in different directions, hence the ≤ and ≥. This makes B automatically our answer. We can check by graphing each of the inequalities on the number line.
x≤5/4
We know that 5/4=1.25. We start at 1.25. Then the inequality tells less than or equal to. Since it is less than, we would travel to the left, where the numbers get smaller and smaller.
x≤5/2
5/2=2.5 We start at 2.5. The inequality tells us greater than or equal to. This means we have to travel to the right, where the number gets larger and larger than 2.5.
Answers
The length of the rectangle is 15 dm and the width is 13 dm.
Let's assume that the length of the rectangle is "L" and the width is "W".
From the problem statement, we have two pieces of information:
The area of the rectangle is 195 dm²:
Area = Length x Width
195 dm² = L x W
The width is two less than the length:
W = L - 2
Now, we can substitute the second equation into the first equation to eliminate W and get an equation with only one variable:
195 dm² = L x (L - 2)
Simplifying the equation:
195 dm² = L² - 2L
L² - 2L - 195 dm² = 0
To solve for L, we can use the quadratic formula:
L = (-b ± √(b² - 4ac)) / 2a
Where a = 1, b = -2, and c = -195.
L = (2 ± √(2² + 4 x 1 x 195)) / 2 x 1
L = (2 ± √4 + 780) / 2
L = (2 ± √784) / 2
L = (2 ± 28) / 2
L = 15 or L = -13
Since the length can't be negative, the length of the rectangle is L = 15 dm.
Now we can use the equation W = L - 2 to find the width:
W = 15 dm - 2 dm
W = 13 dm
Therefore, the length of the rectangle is 15 dm and the width is 13 dm.
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