Let, width = x
Length = x + 6
Now, perimeter, 2(l+b) = 2(x+6+x)
2(2x+6) = 32
4x + 12 = 32
4x = 20
x = 5 &
x+6 = 5+6 = 11
As rectangle's dimensions are 11*5, it's area would be: 11 * 5 = 55 yards²
So, your final answer is 55 Yd²
Hope this helps!
Answer:
The answer is No Solutions
Step-by-step explanation:
Since this is absolute value, we know the answer has to be positive, because distance has to be positive. We see it is negative, and that cannot be, so the answer is no solutions.
Answer:
Step-by-step explanation:
Any time you have compounding more than once a year (which is annually), unless we are talking about compounding continuously, you will use the formula

Here's what we have:
The amount after a certain time that she has in the bank is 4672.12; that's A(t).
The interest rate in decimal form is .18; that's r.
The number of times the interest compounds is 12; that's n
and the time that the money is invested is 3.5 years; that's t.
Filling all that into the formula:
Simplifying it down a bit:
Raise 1.015 to the 42nd power to get
4672.12 = P(1.868847115) and divide to get P alone:
P = 2500.00
She invested $2500.00 initially.
Answer:
x>3
Step-by-step explanation:
Answer:

Step-by-step explanation:
We can rewrite the equation as

Notice that we have
in both the numerator and the denominator, so it looks like we can divide it out. However, what if
is
? Then we would have
, which is undefined. So although it looks like the numerator and denominator can be simplified, the resulting function we would get from simplification would not have the same behavior as this one (since such a function would be defined for
, but this one is not).
A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at
. In the simplified version of the function, if we plug in
, we get

So the point
is our only point of discontinuity.
It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at
(since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of 