<span>This is asking for a linear function. The base pay is $150, so even if she sells nothing, she cannot be paid less than $150. Because of this, we make that out y-intercept. The number of appliances sold is a variable, so we make that the "X" and multiply that by the amount made ($45) on each additional item sold. You then follow the Order of Operations (PEMDAS) to make your equation.</span>
To solve this, you need to isolate/get the variable "x" by itself in the inequality:
2(1 - x) > 2x Divide 2 on both sides

1 - x > x Add x on both sides to get "x" on one side of the inequality
1 - x + x > x + x
1 > 2x Divide 2 on both sides to get "x" by itself
or
(x is any number less than 1/2)
[Another way you could've solved it]
2(1 - x) > 2x Distribute 2 into (1 - x)
(2)1 + (2)(-x) > 2x
2 - 2x > 2x Add 2x on both sides
2 - 2x + 2x > 2x + 2x
2 > 4x Divide 4 on both sides to get "x" by itself


Answer:
35
Step-by-step explanation:
We know the factors of Lena's age are 2 and 5. The least common multiple must have these factors and the factors of 14, so will at least have factors of 2, 5, and 7.
Apparently, the dad's age is 5·7 = 35.
___
The GCF is 5; the LCM is 70 = 5×14.
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Sometimes, I use a little 3-part diagram to think about LCM and GCF. Here, it would look like ...
(2 [5) 7]
where the numbers in curved brackets (2·5) and the numbers in square brackets [5·7] are factors of the two numbers of concern (Lena's age, her dad's age). The middle number in both brackets [5) is the greatest common factor, and the product of all three numbers is their least common multiple.
Here, the product of outside numbers, 2·7 = 14, represents the ratio of the LCM to the GCF. We know that Lena's age has factors of only 2 and 5, so the numbers in the diagram have to be (2[5)7], where 2 and 7 are on the ends and 5 is in the middle.
Answer:
y =
Step-by-step explanation:
the graph shows the vertex is (2,0); which means it is
y =
which show 2 units to the right on the x-axis and staying at y=0
Answer:
Step 1 should have a -2 substituted for both values.