x f(x)–3–1 0 2 5
x g(x)–3–1 0 2 5
For what value of the what value of the domain {–3, –1, 0, 2, 5} does f(x) = g(x) {–3, –1, 0, 2, 5} does f(x) = g(x)? Answer:
consider the relation {(–4, 3), (–1, 0), (0, –2), (2, 1), (4, 3)}.Graph the relation.
State the domain of the relation. State the range of the relation. Is the relation a function? How do you know? Answer:
2. graph the function f(x) = |x + 2|.
Answer:
consider the following expression. Rewrite the expression so that the first denominator is in factored form. Determine the LCD. (Write it in factored form.) Rewrite the expression so that both fractions are written with the LCD. Subtract and simplify.
Answer:
Answer:
-11 and 0 for EDGE2020
f(4)= -11
If g(x)=2, x= 0
Step-by-step explanation:
Answer: 55 can not be considered a significantly high number of girls.
The probability is relevant to answering that question = .
Step-by-step explanation:
Given : For 100 births,
and .
Also it is said that consider a number of girls to be significantly high if the appropriate probability is 0.05 or less
To check :- If 55 girls in 100 births a significantly high number of girls .
We can see that
and , it means it is likely to occur 55 or more girls .
So 55 can not be considered a significantly high number of girls.
The correct answer is:
The numbers are 14 and 16.
Explanation:
Let x and y represent the numbers. Since the sum of the numbers is 30, this gives us the equation
x+y = 30.
Since the difference of the numbers is 2, this gives us the equation
x-y = 2.
This gives us the system
To solve this, we will eliminate one variable. Since the coefficients are all the same, but the y-variables have different signs, we will eliminate them by adding the equations together:
Divide both sides by 2:
2x/2 = 32/2
x = 16.
Substitute this back into our first equation:
16+y=30
Subtract 16 from each side:
16+y-16=30-16
y=14
tan ⊖ + cot ⊖ = 1/sin⊖cos⊖