Median is where if you put them in ascending order and cross them off, one from each side at a time, you get to the middle number.
56789 is your answer but so would any group of 5 numbers where two are bigger and two smaller than seven.
4y-8>10
Add 8
4y-8+8>10+8
4y>18
Now divide both sides by 4
4y/4 > 18/4
y>9/2
Now let's see if it will be true
Replace y by its value
4(9/2)-8>10
36/2 -8 > 10
18-8 > 10
10 > 10
That's false because 10 can not smaller than 10
so it should be = instead
Answer : False
I hope that's help:)
Answer:
- (6-u)/(2+u)
- 8/(u+2) -1
- -u/(u+2) +6/(u+2)
Step-by-step explanation:
There are a few ways you can write the equivalent of this.
1) Distribute the minus sign. The starting numerator is -(u-6). After you distribute the minus sign, you get -u+6. You can leave it like that, so that your equivalent form is ...
(-u+6)/(u+2)
Or, you can rearrange the terms so the leading coefficient is positive:
(6 -u)/(u +2)
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2) You can perform the division and express the result as a quotient and a remainder. Once again, you can choose to make the leading coefficient positive or not.
-(u -6)/(u +2) = (-(u +2)-8)/(u +2) = -(u+2)/(u+2) +8/(u+2) = -1 + 8/(u+2)
or
8/(u+2) -1
Of course, anywhere along the chain of equal signs the expressions are equivalent.
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3) You can separate the numerator terms, expressing each over the denominator:
(-u +6)/(u+2) = -u/(u+2) +6/(u+2)
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4) You can also multiply numerator and denominator by some constant, say 3:
-(3u -18)/(3u +6)
You could do the same thing with a variable, as long as you restrict the variable to be non-zero. Or, you could use a non-zero expression, such as 1+x^2:
(1+x^2)(6 -u)/((1+x^2)(u+2))
The picture isn’t clear enough
1. The first 10 multiples of 13 are 13, 26, 39, 52, 65, 78, 91, 104, 117, and 130.
2. x - 5
3. x + 80
