The range is 327 the maxim subtract the minimum
Answer:
the photo for me is all blurry
Step-by-step explanation:
Answer:
![-5, 18, \sqrt{13}](https://tex.z-dn.net/?f=-5%2C%2018%2C%20%5Csqrt%7B13%7D)
Step-by-step explanation:
We can solve the first equation, f of -3. The value of the function f is
, and plugging in -3 gets us
, this results in 10 divided by negative 2, which is negative 5.
Now, we must solve g of negative one third. The function g is defined as
. Plugging in negative one third into the question gets us ![|9(-\frac{1}{3})-15|](https://tex.z-dn.net/?f=%7C9%28-%5Cfrac%7B1%7D%7B3%7D%29-15%7C)
9 times negative one third is -3, and -3 minus 15 is -18. The absolute value of -18 is 18.
Now, we must solve h of negative 2, and h is defined as
. Plugging in negative 2, we have
. Negative 8 times negative 2 is positive 16, and 16 minus 3 is 13. The answer is the square root of 13
Answer:
Data Set 1 and 3.
Step-by-step explanation:
For a table to be a function, there cannot be multiple x values of the same number. This means that a function must be one-to-one.
When looking at the data sets, we can eliminate them by using this logic. Data Set 2 contains two y values at y = 6, as well as two x values at x = 1. Therefore, this is not a function.
Data Set 1 is the only table that represents a function because it consists of different x values without any repeating.
Data Set 3 also contains different x values without repetition. Each x value has a specific y value. This is a function as well.
=±22
x
=
±
2
x
2
Using the fact that 2=ln2
2
=
e
ln
2
:
=±ln22
x
=
±
e
x
ln
2
2
−ln22=±1
x
e
−
x
ln
2
2
=
±
1
−ln22−ln22=∓ln22
−
x
ln
2
2
e
−
x
ln
2
2
=
∓
ln
2
2
Here we can apply a function known as the Lambert W function. If =
x
e
x
=
a
, then =()
x
=
W
(
a
)
.
−ln22=(∓ln22)
−
x
ln
2
2
=
W
(
∓
ln
2
2
)
=−2(∓ln22)ln2
x
=
−
2
W
(
∓
ln
2
2
)
ln
2
For negative values of
x
, ()
W
(
x
)
has 2 real solutions for −−1<<0
−
e
−
1
<
x
<
0
.
−ln22
−
ln
2
2
satisfies that condition, so we have 3 real solutions overall. One real solution for the positive input, and 2 real solutions for the negative input.
I used python to calculate the values. The dps property is the level of decimal precision, because the mpmath library does arbitrary precision math. For the 3rd output line, the -1 parameter gives us the second real solution for small negative inputs. If you are interested in complex solutions, you can change that second parameter to other integer values. 0 is the default number for that parameter.