Answer:
The right choices are
Point B is plotted at negative one-half.
Point D is a rational number.
Point F is plotted at three-fourths.
Step-by-step explanation:
Let's make a number line according to the given situation. That would help us to answer the problem.
From the diagram,
Point B is plotted at negative one-half.
Point D is a rational number.
Point F is plotted at three-fourths are correct!
Answer:
Real numbers for both
Step-by-step explanation:
The domain of a function is the set of values that the unknown t can adopt. For this function, t can be any real number as there are no restrictions for the t. Ir can be any positive number, 0, negative numbers, fractions, irrational numbers, whatever number you like.
The range of a function is the values that p(t) adopt when we replace the t value with any number. Here, again, it range is all real numbers. If you want p(t) to be positive it is possible, negative is possible, 0 is possible, and so on. If you like, you can verify it by replacing the numbers you like.
Something to know is that linear polynomial functions ALWAYS have their domains and ranges in real numbers.
5 x (4 x 3) = (5 x 4) x 3
7 x (3 x 6) = (7 x 3) x 6
(30 x 5) x 12 = (30 x 12) x 5
first half
<u>Option A)</u> when y=6 and w= 2, <u>Option B)</u> when y=4 and w=-1 are the TWO correct choices.
Step-by-step explanation:
The given inequality is w<x<y and x>2.
<u>step 1</u>: let us <u>assume that x=3</u>, since it is given that x>2.
<u>step 2</u>: substitute the value of x=3 and each options in the inequality
<u>step 3</u>: option A) w=2, y=6. Then w<x<y = 2<3<6 (the condition satisfies).
option B) w= -1, y=4. Then w<x<y = -1<3<4 (the condition satisfies).
option C) w= 0, y=2. then w<x<y ≠ 0<2<2 (does not satisfies).
option D) w= -2, y=0. then w<x<y ≠ -2<2<0 (does not satisfies).
option E) w= -4, y= -1. then w<x<y ≠ -4<2<-1 (does not satisfies).
