Multiple the bionomials and then PICK YOUR ANSWER FROM THE ANSWER BANK!

Liono4ka [1.6K]

(if 2 is a square)

6xy ( x2 - xy + y2)

= 6x3y - 6x2y2 - 6xy3 --------------the threes and twos on this line are squares and cubics. the six is just a whole number

6 0

3 years ago

The function f(x) is given by the set of ordered pairs. {(1,0), (–10,2), (0,6), (3,17), (–2,–1)} Which equation is true? f of ne

Rudik [331]

Answer:

f(0)=6

Step-by-step explanation:

When representing function's values by set of ordered pairs, the first one correspond to the x value and the second to the image or "y" value.

When you read f(0) means "when x is 0" then the 6 is the value of the function or the image.

7 0

2 years ago

Read 2 more answers

Last questions, plz help

Goryan [66]

Answer:

9. $a_n=7+6n$

10. $a_n=a_{n-1}+7$

Step-by-step explanation:

9. Given the sequence

$13, 19, 25, ...$

In this sequence,

$a_1=13\\ \\a_2=19\\ \\a_3=25\\ \\...$

Note that

$a_2-a_1=19-13=6\\ \\a_3-a_2=25-19=6,$

then the common difference in this sequence is $d=6$

You have

$a_1=13\\ \\d=6,$

then the explicit formula is

$a_n=a_1+(n-1)d\\ \\a_n=13+(n-1)\cdot 6\\ \\a_n=13+6n-6\\ \\a_n=7+6n$

10. Given the sequence

$-10,-3, 4, 11, ...$

In this sequence,

$a_1=-10\\ \\a_2=-3\\ \\a_3=4\\ \\a_4=11\\ \\...$

Note that

$a_2-a_1=-10-(-3)=-10+3=-7\\ \\a_3-a_2=-3-4=-7\\ \\a_4-a_3=11-4=7,$

then the common difference in this sequence is $d=7$

You have

$a_1=-10\\ \\d=7,$

then the recursive formula is

$a_n=a_{n-1}+d\\ \\a_n=a_{n-1}+7$

5 0

3 years ago

In a group of 60 numbers, average of

Arada [10]

Answer:

c. 18

Step-by-step explanation:

First, you would find the total sum of the 60 numbers. The average is always the sum of the numbers divided by the number of numbers.

35*18=630

15*10=150

Now we do not know how many numbers the last group has. However, we can find it out doing 60-35-15 to see how many numbers are left. There would be 10 numbers left

10*30=300

630+150+300=1080

1080/60=18

Hence, the answer is C(18)

4 0

3 years ago

1) AU (C - B) =

torisob [31]

Answer:

$A \cup (C-B) = (A \cup C) \cap (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C = A \cup (B \cup C)\\B^c \cap (A-C) = (B^c - C) \cap A\\(B \cap C) - A = (B-A)\cap C\\C-(A\cup B) = (C - A )\cap B\\A^c \cap (B \cap C) = ( A^c \cap B) \cap C$

Step-by-step explanation:

$A \cup (C-B) = A \cup (C \cap B^c) = (A \cup C) \cap (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C = (A \cup B ) \cup C = A \cup (B \cup C)\\B^c \cap (A-C) = B^c \cap (A \cap C^c) = (B^c \cap C^c) \cap A = (B^c - C) \cap A\\(B \cap C) - A = (B \cap C) \cap A^c = (B \cap C) \cap A^c = (B \cap A^c)\cap C = (B-A)\cap C\\C-(A\cup B) = C \cap (A\cup B)^c = C \cap (A^c \cap B^c ) = (C \cap A^c )\cap B^c = (C - A )\cap B\\$

$A^c \cap (B \cap C) = ( A^c \cap B) \cap C$

3 0

3 years ago