Complete the table for the equation of the line y = 3x - 1 X
f(x)
-1
0
2
Answers
Answer:
Step-by-step explanation:
Given the equation of a line expressed as y = 3x-1, we are to find y = f(x) for the following values of x in the table.
If y = 3x-1, then f(x) = 3
When x = -1
f(-1) = 3(-1)-1
f(-1) = -3-1
f(-1) = -4
Hence when x = -1, f(x) = -4
When x = 0
f(0) = 3(0)-1
f(0) = 0-1
f(0) = -1
Hence when x = 0, f(x) = -1
When x = 2
f(2) = 3(2)-1
f(2) = 6-1
f(2) = 5
Hence when x = 2, f(x) = 5
Final answer:
To complete the table for the equation of the line, substitute the given x-values into the equation and solve for y.
Explanation:
To complete the table for the equation of the line y = 3x - 1, we substitute the given x-values into the equation and solve for y. For x = -1, y = 3(-1) - 1 = -4. For x = 0, y = 3(0) - 1 = -1. And for x = 2, y = 3(2) - 1 = 5. Therefore, the completed table is:
x
f(x)
-1
-4
0
-1
2
5
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For me, I just did 15 divided by 3, which equals to 5. So.. You do 5 x 5 to find out how many people there are in the club in total.
Final answer:
The question is about finding the total number of people in the yearbook club given that 15 people represent three-fifths of the total. By using a proportion, we find that the yearbook club has 25 people.
Explanation:
This is a proportional relationship problem in mathematics. You are given that 15 people represent three-fifths (or 3/5) of the total number of people in the yearbook club. To find the total number of people in the club, you set up the proportion: 15 is to 3 and X is to 5.
Then, cross multiply and solve the equation for X. 3*X = 15*5, therefore X = 75/3 = 25. So, there are 25 people in the yearbook club.
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Answers
Answer:
and
Step-by-step explanation:
we need to write the fraction in two different ways.
we write two equivalent fraction for the given fraction
Divide the given fraction by 3 on both sides top and bottom
To get another equivalent fraction , multiply the given fraction by any number at the top and bottom. Multiply the fraction by 2
To fractions are and
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ln((5x-1)(x-6)=ln6 do now FOIL
ln(5x^2-30x-x+6)=ln6
therefore
5x^2-31x+6=6 and
5x^2-31x=0 furthermore
x(5x-31)=0
one solution is 0, which we will reject
the other solution is 5x=31
x=31/5
ln((5x - 1)(x - 6)) = ln(6)
ln(5x(x - 6) - 1(x - 6) = ln(6)
ln(5x(x) - 5x(6) - 1(x) - 1(-6)) = ln(6)
ln(5x² - 30x - x + 6) = lm(6)
ln(5x² - 31x + 6) = ln(6)
5x² - 31x + 6 = 6
5x² - 31x + 6 - 6 = 6 - 6
5x² - 31x + 0 = 0
x = -(-31) +/- √((-31)² - 4(5)(0))
2(5)
x = 31 +/- √(961 - 0)
10
x = 31 +/- √(961)
10
x = 31 +/- 31
10
x = 31 + 31 x = 31 - 31
10 10
x = 62 x = 0
10 10
x = 6²/₅ x = 0