So I'm not 100% on this, but I believe that first you need to combine like terms. The 4x and the -2x make 2x and distribute the 2 so you get 2x+1. That all equals 2x+5=2x+2. You need to get the variable by themselves so I will minus the 2. You then get 2x+3=2x. You then need to minus the 2x, which makes 0. So the answer is 3=0, or undefined. Is that about what you guys are working on? Have you learned about undefined?
Answer:
I will be answering numbers 3), 4), and 6) (so that you get most of the concept and that you are able to solve the questions by yourself), and also 9).
Step-by-step explanation:
3)
-15/4b+5/6b
So, first find the common denominator of both fractions. 4 times 6 is 24. -15/4b= -90/24b. 5/6b=20/24b. -90/24b+20/24b=-70/24b=-35/24b=-1 11/24.
4)
60m-15(4-8m)+20
What we need to do first is remove the parentheses.
60m-(15*4-15*8m)+20
=60m-60-120m+20.
So, here, we add the numbers without 'm' and the numbers with 'm.' (Always include the sign in front of the number, like in the equation, we include -60 since that's the sign in front of 60.)
60m-60-120m+20
=-60+20=-40
60m-120m=-60m
=-60m-40
6)
9y-15y+12-6y
Like I said before on number 4, we do about the exact same thing (but there are no parentheses to remove, so we skip that step.)
9y-15y+12-6y
=9y-15y-6y
=-12y+12
9)
It is basically the same thing as we did in the previous problems, just in a different way of saying it.
12x+12x+2x+2x+x+9
=28x+9 yards
Answer:
9.42 cubic inches
Step-by-step explanation:
In order to find the volume for a sphere, we use the formula V = πr. For the sake of simplicity, we will be using 3.14 for π. Before we plug anything in, let's get our radius. Remember, the radius is half of the diameter.
Equation: d = 2r
Replace: 6 = 2r
Divide: 3 = r
Now that we have our radius, let's plug in what we have and solve.
Equation: V = 3.14r
Replace: V = 3.14(3)
Multiply: V = 9.42
Remember, volume is measured in cubic units, so use cubic inches for the unit here.
The cost of 5 apples is $6
Answer:
a) there is s such that <u>r>s</u> and s is <u>positive</u>
b) For any <u>r>0</u> , <u>there exists s>0</u> such that s<r
Step-by-step explanation:
a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.
b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.
