Answer:
Choice B is correct; the domain of function A is the set of real numbers greater than 0
The domain of the function B is the set of real numbers greater than or equal to 1
Step-by-step explanation:
The domain of a function refers to the set of x-values for which the function is real and defined. The graph of function B reveals that the function is defined when x is equal 1 and beyond; that is its domain is the set of real numbers greater than or equal to 1.
On the other hand, the natural logarithm function is defined everywhere on the real line except when x =0; this will imply that its domain is the set of real numbers greater than 0 . In fact, the y-axis or the line x =0 is a vertical asymptote of the natural log function; meaning that its graph approaches this line indefinitely but neither touches nor crosses it.
Answer:
Step-by-step explanation:
Hello!
The five-number summary comprehends the minimum and maximum values, the 1st and 3rd quartiles and the median.
1st step: is to arrange the data sets from least to greatest.
Sherelle:
26 39 56 58 60 62 65 66 66 68 71 72 72 73 74 75 81 83 84 85
Venita:
44 45 51 51 53 53 55 57 58 62 65 66 69 69 70 73 75 77 78 79
2nd step: To calculate each quartile you have to determine their position and then identify which observation sits in that position:
Quartile 1:
PosQ₁: n/4= 20/4= 5
Quartile 2 (Median)
PosMe: n/2= 20/2= 10
Quartile 3
PosQ₃: n*(3/4)= 20*(3/4)= 15
Since both samples have the same size, the 1st quartile will be the fifth observation of each sample, the median will be the tenth observation and the 3rd quartile will be the fifteenth.
Sherelle:
Q₁: 60
Me: 68
Q₃: 74
Minimum: 26
Maximum: 85
Venita:
Q₁: 53
Me: 62
Q₃: 70
Minimum: 44
Maximum: 79
I hope this helps!
Answer:
h = -6
Step-by-step explanation:
Use the quadratic formula
ℎ
=
−
±
2
−
4
√
2
h=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
h=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
ℎ
2
−
3
6
=
0
h^{2}-36=0
h2−36=0
=
1
a={\color{#c92786}{1}}
a=1
=
0
b={\color{#e8710a}{0}}
b=0
=
−
3
6
c={\color{#129eaf}{-36}}
c=−36
ℎ
=
−
0
±
0
2
−
4
⋅
1
(
−
3
6
)
(5-4)-3 = -2. If we move the parenthesis, as is defined by what the associative property is, 5-(4-3) = 4. No matter what, you always approach mathematical equations or expressions using PEMDA. We have to do what's inside the parenthesis first, which is why we end up with 2 different solutions using subtraction.
