All categories

3 years ago

Can someone help me with my work

Mathematics

1 answer:

juin [17]3 years ago

5 0

F(x) = x
stretch by 3:
f(x) = 3x

flip over x-axis
g(x) = -3x

answer is C. g(x) = -3x

You might be interested in

2x+8y=54<br> x+9y=67<br> Solve the system by subtitution

Kisachek [45]

Step-by-step explanation:

2x +8y = 54

x + 9y = 67

x = 67 - 9y

2(67-9y) + 8y = 54

434 - 18y + 8y = 54

434 - 10y = 54

10y = 54 - 434

10y = -380

y = -38

3 0

3 years ago

Elise wrote all possible four-digit numbers using two of the digit 2 and two of the digit 0. Two of Elise’s numbers have a great

Lyrx [107]

Answer:

The required difference is 198.

Step-by-step explanation:

The four digit smallest number using 2,2,0,0 is =2002

The four digit largest number using 2,2,0,0 is =2200

Therefore the difference between 2200 and 2002 is =198

6 0

3 years ago

Lamar wants to replace a glass window in his restaurant. The window is in the shape of a square. Its side lengths are 11 feet. S

Over [174]

Answer:

55x11 = 55

(Meeting character limit)

7 0

2 years ago

Read 2 more answers

The data in which table represents a linear function that has a slope of zero? A 2-column table with 5 rows. Column 1 is labeled

lisov135 [29]

Answer: A 2-column table with 5 rows. Column 1 is labeled x with entries negative 5, negative 4, negative 3, negative 2, negative 1. Column 2 is labeled y with entries 5, 5, 5, 5, 5.

8 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago