now, exponential functions have a horizontal asymptote at the x-axis, namely when y = 0, however, if you just move this one with a vertical translation of 2, then the horizontal asymptote will be at 2 instead.
y = 3(2)ˣ + 2
Patrick worked 160.5 hours because you have to subtract 1000 from 7420 then divide that by 40 for your answer. Also, Patrick has a sweet job.
The given statement is:
An integer is divisible by 100 if and only if its last two digits are zeros
The two conditional statements that can be made are:
1) If an integer is divisible by 100 its last two digits are zeros.
This is a true statement. If a number is divisible by 100, it means 100 must be a factor of that number. When 100 will be multiplied by the remaining factors, the number will have last two digits zeros.
2) If the last two digits of an integer are zeros, it is divisible by 100.
This is also true. If last two digits are zeros, this means 100 is a factor of the integer. So the number will be divisible by 100.
Therefore, the two conditional statements that are formed are both true.
So, the option A is the correct answer.
Yes, it is. When the definition is separated into two conditional statements, both of the statements are true
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))
