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2 years ago

The function f(x) = x2 has been translated 9 units up and 4 units to the right to form the function g(x). Which represents g(x)?

Mathematics

2 answers:

tester [92]2 years ago

8 0

Hello the answer is

c)g(x) = (x − 4)2 + 9

Alenkinab [10]2 years ago

7 0

For this case, the parent function is given by:

$f (x) = x ^ 2$

We apply the following transformations:

Vertical translations:

Suppose that k> 0

To graph y = f (x) + k, move the graph of k units upwards:

For k = 9 we have:

$h (x) = x ^ 2 + 9$

Horizontal translations:

Suppose that h> 0

To graph y = f (x-h), move the graph of h units to the right

For h = 4 we have:

$g (x) = (x-4) ^ 2 + 9$

Answer:

The function g (x) is given by:

$g (x) = (x-4) ^ 2 + 9$

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Answer:

Y = 6

Step-by-step explanation:

On the first day of ticket sales the school sold 3 adult ticket and 8 student tickets for a total of $72. The school took in $152 on the second day by selling 7 adult tickets and 16 student tickets. How much is a student ticket?

Day 1: 3x + 7y = 72

Day 2: 7x + 16y = 152

This becomes a system of equations.

3x+8y=72

3x+8y+−8y=72+−8y (Add -8y to both sides)

3x=−8y+72

3x/3 = −(8y+72)/3

X = (-8/3) y + 24

Substitute (-8/3) y + 24 for x in7x+16y=152:

7(-8/3 y + 24) + 16y = 152

-8/3 y + 168 = 152 (Simplify both sides of the equation)

-8/3 y + 168 − 168 = 152 + (−168) (Add -168 to both sides)

-8/3 y = -16

(-8/3 y)/(-8/3 y) = -16/ -8/3 y (Divide both sides by (-8)/3)

y = 6

If you wanted to keep going...

Substitute 6f or y in x = (-8/3) y + 24

(-8/3) y + 24 = (-8/3) (6) + 24

SImplify: x = 8

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Answer:

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Answer:

3. What is the probability that an adult selected at random has both a landline and a cell phone?

A. 0.58

4. Given an adult has a cell phone, what is the probability he does not have a landline?

C. 0.3012

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that an adult has a landline at his residence.

B is the probability that an adult has a cell phone.

C is the probability that a mean is neither of those.

We have that:

$A = a + (A \cap B)$

In which a is the probability that an adult has a landline but not a cell phone and $A \cap B$ is the probability that an adult has both of these things.

By the same logic, we have that:

$B = b + (A \cap B)$

The sum of all the subsets is 1:

$a + b + (A \cap B) + C = 1$

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This means that $C = 0.02$ .

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So $A = 0.73, B = 0.83$ .

What is the probability that an adult selected at random has both a landline and a cell phone?

This is $A \cap B$ .

We have that $A = 0.73$ . So

$A = a + (A \cap B)$

$a = 0.73 - (A \cap B)$

By the same logic, we have that:

$b = 0.83 - (A \cap B)$ .

$a + b + (A \cap B) + C = 1$

$0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1$

$(A \cap B) = 0.75 + 0.83 - 1 = 0.58$

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

$b = B - (A \cap B) = 0.83 - 0.58 = 0.25$

25% of those do not have a landline.

So $P = \frac{0.25}{0.83} = 0.3012$

The answer for question 4 is C.

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