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Answer:
Problem 2): 
which agrees with answer C listed.
Problem 3) : D = (-3, 6] and R = [-5, 7]
which agrees with answer D listed
Step-by-step explanation:
Problem 2)
The Domain is the set of real numbers in which the function (given by a graph in this case) is defined. We see from the graph that the line is defined for all x values between 0 and 240. Such set, expressed in "set builder notation" is:

Problem 3)
notice that the function contains information on the end points to specify which end-point should be included and which one should not. The one on the left (for x = -3 is an open dot, indicating that it should not be included in the function's definition, therefor the Domain starts at values of x strictly larger than -3. So we use the "parenthesis" delimiter in the interval notation for this end-point. On the other hand, the end point on the right is a solid dot, indicating that it should be included in the function's definition, then we use the "square bracket notation for that end-point when writing the Domain set in interval notation:
Domain = (-3, 6]
For the Range (the set of all those y-values connected to points in the Domain) we use the interval notation form:
Range = [-5, 7]
since there minimum y-value observed for the function is at -5 , and the maximum is at 7, with a continuum in between.
Y⁴ + 12y² + 36
Now factorize the expression
y⁴ + 6y² + 6y² + 36
= y²(y² + 6) + 6(y² + 6)
= (y² + 6) (y² + 6)
<span>Now 6 is not the perfect square and according to rule, binomial can not be factored as the difference of two perfect squares.
</span>so multiply both.
(y² + 6)² is the answer.
Step-by-step explanation:
A) 0.27
B) 1/4 = 0.25
C) 3/8 = 0.375
D) 2/ 11 = 0.1818
E) 11% = 0.11
E is the correct answer
Answer:
When point A with coordinates (0, -1) is reflected across the x-axis and mapped onto point A', the coordinates of A' will be (0, 1).
i.e A'(0, 1) is the image of point A after a reflection.
Hence, point A is reflected across the x-axis.
Step-by-step explanation:
When we reflect a point A across the x-axis, the value of 'y' gets negated, but the value of 'x' remains unchanged.
In other words, when point P with coordinates (x, y) is reflected across the x-axis and mapped onto point P', the coordinates of P' will be (x, -y).
Thus, the rule is:
P(x, y) → P'(x, -y)
Thus, when point A with coordinates (0, -1) is reflected across the x-axis and mapped onto point A', the coordinates of A' will be (0, 1).
i.e A'(0, 1) is the image of point A after a reflection.
Hence, point A is reflected across the x-axis.