a) because the denominators are the same add the numerators:
1 1/5 + 3 2/5 = 4 3/5
b) rewrite the fractions to have a common denominator:
1/2 = 3/6
1/3 = 2/6
Now subtract:
4 3/6 - 1 2/6 = 3 1/6
Answer:
Set of all real numbers
Step-by-step explanation:
When we talk of domain, we mean the possible values of x that the equation can take without being undefined
So x-values that makes the equation defined are the domain of the graph
From the diagram, we can see that the graph extends from negative infinity to positive infinity
What this mean is that the possible values here come from the top right and extends to the left
So this means that the range of possible values are the set of all real numbers
We know A=L•W so set 43 equal to A. Equation should look like this thus far: 43=L•W. Now when the question says more than we know that means adding. Since we don’t know the width let w=width. So know your equation should look like this: 43=2+w•(w). Knowing this equation you should be able to solve for the dimensions of the sign. If you need help on that let me know.
Answer:
A) x = 3 or -1
B) x = -7
C)x = -7
Step-by-step explanation:
A) x² + 2x + 1 = 2x² - 2
Rearranging, we have;
2x² - x² - 2x - 2 - 1 = 0
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
B) ((x + 2)/3) - 2/15 = (x - 2)/5
Multiply through by 15 to get;
5(x + 2) - 2 = 3(x - 2)
5x + 10 - 2 = 3x - 6
5x - 3x = -6 - 10 + 2
2x = -14
x = -14/2
x = -7
C) log(2x + 3) = 2log x
From log derivations, 2 log x is same as log x²
Thus;
log(2x + 3) = logx²
Log will cancel out to give;
2x + 3 = x²
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
