I need a-n answers!!! help plseaseeeee

Mathematics

1 answer:

bekas [8.4K]3 years ago

5 0

A = consecutive interior
b= alternate exterior
c= consecutive exterior
d= alternate interior
e= corresponding
f= none
g= consecutive exterior
h= corresponding
i= alternate exterior
j= consecutive exterior
k=corresponding
l= alternate exterior
m= alternate interior
n= consecutive interior

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What does x equal in x^3-9x+1=0

12345 [234]

X^3 -9x+1= 0 (minus 1)

X^3-9x=-1 (divide by 9)

x^3-x=-1/9

But it doesn’t work any further so you have written it wrong

5 0

3 years ago

The graph of the parent function y = x cubed is horizontally stretched by a factor of One-fifth and reflected over the y-axis. W

sasho [114]

Answer: OPTION D.

Step-by-step explanation:

Below are some transformations for a function f(x):

If $-f(x)$ , then the function is reflected over the x-axis.

If $f(-x)$ , then the function is reflected over the y-axis.

If $f(bx)$ and $b>1$ , then the function is horizontally compressed.

If $f(bx)$ and $0 < b$ , then the function is horizontally stretched.

In this case you know that the parent function is:

$y=x^3$

According to the information given in the exercise, the parent function IS horizontally stretched by a factor of $\frac{1}{5}$ and it is also reflected over the y-axis.

Therefore, based on the transformations explained before, you can notice that the transformation is:

$f(-bx)$

Where $0 < b$

Therefore, the equation of the transformed function is:

$y=(-\frac{1}{5}x)^ 3$

8 0

3 years ago

Read 2 more answers

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$