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Licemer1 [7]
3 years ago
11

Which of the following is an arithmetic sequence?

Mathematics
1 answer:
ira [324]3 years ago
4 0

Answer:

B- because you add 5 eah time

Step-by-step explanation:

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Please help me round 34,699 to the nearest ten thousand
Serhud [2]
Okay so rounding to the ten thousand will mean you are looking at the 3 and looking behind it, the 4 is behind the 3 in 34,699 and 0-4 we round down so the nearest 10,000 is 30,000. ANSWER: 30,000
4 0
3 years ago
Can some one help me. I don’t understand
vovikov84 [41]
BC = √[(8-1)^2 +(1-6)^2]
BC = √(49+25
BC = √74
BC = 8.6

answer
8.6 units
4 0
4 years ago
Read 2 more answers
Help pls
timurjin [86]

Answer:

s> 144 2/3

Step-by-step explanation:

I took the test in k12 and i got it right so yea !

3 0
2 years ago
Expand the given power by using Pascal’s triangle. (9a - 10b)^6
xxMikexx [17]

Answer:

531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6

Step-by-step explanation:

                     1                                n=0

                  1      1                            n=1

                 1   2   1                           n=2

                1  3  3  1                          n=3

              1  4  6  4   1                      n=4

             1  5 10 10 5  1                    n=5

           1 6 15 20 15 6 1                   n=6

This is where n is the exponent in

(x+y)^n.

(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6

Now we want to expand:

(9a-10b)^6 or we we can rewrite as (9a+(-10b))^6.

Let's replace x with (9a) and y with (-10b) in the expansion:

(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6

((9a)+(-10b))^6

=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6

Let's simplify a bit:

=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6

=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6

8 0
4 years ago
Find the inverse of the function.
asambeis [7]

Answer:

A) f^{-1}(n)= (n+2)^3

Step-by-step explanation:

f(n) = \sqrt[3]{n} -2

y = \sqrt[3]{n} -2

y = n^{1/3}-2

n = y^{1/3} -2

n + 2 = y^{1/3}

y = (n + 2) ^3

f^{-1}(n) = (n + 2)^3

Hope this helps!

7 0
3 years ago
Read 2 more answers
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