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vredina [299]
1 year ago
6

Which table shows a function that is decreasing only over the interval (–1, ∞)? A 2-column table with 6 rows. The first column i

s labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 1, negative 3, negative 5, negative 2, negative 1, 2. A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 3, negative 5, negative 7, negative 6, 1, negative 1. A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 4, negative 3, negative 1, 2, 1, negative 6. A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 5, negative 1, 1, 0, negative 4, negative 8. Mark this and return
Mathematics
1 answer:
Lemur [1.5K]1 year ago
8 0

Answer:

B. f(x) ≤ 0 over the interval [0, 2]. D.f(x) > 0 over the interval (–2, 0). E.f(x) ≥ 0 over the interval [2, ).

Step-by-step explanation:

Those are the 3 answers. Just did it on edge.

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If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?
kiruha [24]

Answer:

[1,\infty)

Step-by-step explanation:

b(x)=\sqrt{x-4}

a(x)=3x+1

Since we want to know the domain of (b \circ a)(x), let's first consider the domain of the inside function, that is, that of a(x)=3x+1. Every polynomial function has domain all real numbers.

So we can plug anything for function a and get a number back.

Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.

Let's find (b \circ a)(x) and simplify now.

(b \circ a)(x)

b(a(x))

b(3x+1)

\sqrt{(3x+1)-4}

\sqrt{3x+1-4}

\sqrt{3x-3}

Now again we can only square root positive or zero numbers so we want 3x-3 \ge 0.

Let's solve this to find the domain of (b \circ a)(x).

3x-3 \ge 0

Add 3 on both sides:

3x \ge 3

Divide both sides by 3:

x \ge 1

So we want x to be a number greater than or equal to 1.

The option that says this is [1,\infty)

-------------------------------

Give an example why option A fails:

A number in the given set is -2.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(-2)=3(-2)+1=-6+1=-5 and b(-5)=\sqrt{-5-4}=\sqrt{-9} \text{ which is not real}.

Give an example why option B fails:

A number in the given set is 0.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(0)=3(0)+1=0+1=1 and b(1)=\sqrt{1-4}=\sqrt{-3} \text{ which is not real}.

Give an example why option D fails:

While all the numbers in set D work, there are more numbers outside that range of numbers that also work.

A number not in the given set that works is 3.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(3)=3(3)+1=9+1=10 and b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}.

4 0
2 years ago
Read 2 more answers
Write an absolute value equation to satisfy the given solution set shown on a number line.
aivan3 [116]

Answer: IX - 4I ≤ 4

Step-by-step explanation:

In the numer line we can see that our possible values of x are in the range:

0 ≤ x ≤ 8

And we want to find an absolute value equation such that this set is the set of possible solutions.

An example can be:

IX - 4I ≤ 4

To construct this, we first find the midpoint M of our set, in this case is 4.

Then we write:

Ix - MI  ≤ IMI

Notice that i am using the minor and equal sign, this is because the black dots means that the values x = 0 and x = 8 are included, if the dots were empty dots, it would be an open set and we should use the <  > signs.

8 0
2 years ago
Exams are approaching and Helen is allocating time to studying for exams. She feels that with the appropriate amount of studying
svp [43]

Answer: 0.05

Step-by-step explanation:

Let M = Event of getting an A in Marketing class.

S = Event of getting an A in Spanish class,

i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45

Required probability = P(neither M nor S)

= P(M'∩S')

= P(M∪S)'                                 [∵P(A'∩B')=P(A∪B)']

=1- P(M∪S)                               [∵P(A')=1-P(A)]

= 1- (P(M)+P(S)- P(M∩S))   [∵P(A∪B)=P(A)+P(B)-P(A∩B)]

= 1- (0.80+0.60-0.45)

= 1- 0.95

= 0.05

hence, the probability that Helen does not get an A in either class= 0.05

3 0
2 years ago
What is the following quotient? sqrt 6 + sqrt 11 / sqrt 5 +sqrt 3
just olya [345]

Answer:

\frac{\sqrt6 +\sqrt{11}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{30} - \sqrt{18} + \sqrt{55} - \sqrt{33}}{2}

Step-by-step explanation:

Let's translate what you've written in words to an expression with mathematical symbols.

\frac{\sqrt6 +\sqrt{11}}{\sqrt{5} + \sqrt{3}}

To solve this, you need to multiply both the numerator and denominator with the conjugate of the denominator.

\frac{\sqrt6 +\sqrt{11}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{30} - \sqrt{18} + \sqrt{55} - \sqrt{33}}{2}

5 0
2 years ago
Read 2 more answers
If the sample space S is an infinite set, does this necessarily imply that any random variable X defined from S will have an inf
alekssr [168]

Answer: no

Step-by-step explanation: see attachment

4 0
1 year ago
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