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Jlenok [28]
2 years ago
13

How does the range of g (x) = StartFraction 6 Over x EndFraction compare with the range of the parent function f (x) = StartFrac

tion 1 Over x EndFraction?
The range of both f(x) and g(x) is all real numbers
The range of both f(x) and g(x) is all nonzero real numbers
The range of f(x) is all real numbers, the range of g(x) is all real numbers except 6
The range of f(x) is all nonzero real numbers, the range of g(x) is all real numbers except 6
Mathematics
2 answers:
xz_007 [3.2K]2 years ago
8 0

Answer:

B

Step-by-step explanation:

Lerok [7]2 years ago
3 0

Answer: Option: (B) The range of both f(x) and g(x) is all nonzero real numbers is correct.

Step-by-step explanation: ..................Because I said so............................

You might be interested in
Find the first, fourth, and eighth terms of the sequence. A(n) = –3 • 2^n–1
Sphinxa [80]
A(n) = –3 • 2⁽ⁿ⁻¹⁾

for n = 1 ,  A₁ = -3.(2)⁰ = -3
for n = 2 ,  A₂ = -3.(2)¹ = -6
for n = 3 ,  A₃ = -3.(2)² = -12
for n = 4 ,  A₄ = -3.(2)³ = -24
...........................................
for n = 8 ,  A₈ = -3.(2)⁷ = -384


4 0
1 year ago
5y-6=3y-20 how do i solve this problem
rusak2 [61]
Minus 3y both sides
2y-6=-20
add 6 to both sids
2y=-14
divide 2
y=-7
7 0
2 years ago
Read 2 more answers
For each system of equations, drag the true statement about its solution set to the box under the system?
natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

 of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

   coefficients of x and y are the same and the numerical terms

   are the same

- The system of equation has one solution if at least one of the

  coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

  numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0
1 year ago
Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll
JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}

So, the possibility of getting a total of 7 = \frac{6}{36}=\frac{1}{6}

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = \frac{2}{36} = \frac{1}{18}

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = \frac{28}{36} = \frac{7}{9}

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11

\frac{11}{3}+\frac{11}{3}-\frac{77}{9}

\frac{22}{3}-\frac{77}{9}

\frac{66-77}{9}

-\frac{11}{9}

-1.22

Therefore the expected payoff for this game is -$1.22.

4 0
1 year ago
Danny Henry made a waffle on his six-inch-diameter circular griddle using batter containing a half a cup of flour. Using the sam
Mekhanik [1.2K]

Answer:

He'll need 288 cups to make a waffle on his 24 foot diameter circular griddle.

Step-by-step explanation:

In order to find out how much batter Danny needs we first need to compute the area of the first pans, since it is a circular pan their area is given by A = 2*pi*r. So we have:

Area of the first pan = 2*pi*(6/2) = 18.84 square inches

Area of the second pan = 2*pi*(24/2) = 75.36 square foots

We now need to convert these values to be in the same unit, we'll convert from square foots to square inches:

Area of the second pan = 75.36 * (12)^2 = 10851.84 square inches

We can now use a proportion knowing that the batter and the thickness of the waffles are the same. If 0.5 cup of flour can make a waffle on 18.84 square inches then x cup of flour can make a waffle on 10851.85 square inches. Writing this in a mathematical form, we have:

0.5/x = 18.84/10851.84

18.84x = 0.5*10851.85

x = 10851.85*0.5/18.84 =288 cups

7 0
2 years ago
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