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3 years ago

14

Choose the correct classifacation of 4x^4-4x^3+10x^6

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:what’s classification

Step-by-step explanation:

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-8x+54x+140 where x is the number of $1 increases in prices

Alenkinab [10]

(-8×1)+(54×1)+140=186

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3 years ago

Anyone? please help asap i really want to finish this

yan [13]

Q. 3

$b \geqslant 2 \\$

<u>b∈(2,∞)</u>

Q. 4

$r < - 5$

<u>r∈(-∞, -5)</u>

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2 years ago

you and a friend decided to take a road trip to somewhere in Pennsylvania on the weekend. You are traveling at an average speed

Zinaida [17]

62 multiplied by 3 is 186
Add on 31 for the half hour
and you get 217

8 0

3 years ago

Let $M$ be the least common multiple of $1, 2, \ldots, 20$. How many positive divisors does $M$ have

Veronika [31]

Consider the prime factorization of 20!.

$20! = 20 \times 19 \times 18 \times \cdots \times 3 \times 2 \times 1$

The LCM of 1, 2, ..., 20 must contain all the primes less than 20 in its factorization, so

$M = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times m$

where $m$ is some integer not divisible by any of these primes.

Compare the factorizations of the remaining divisors of 20!, and check off any whose factorizations are already contained in the product of primes above.

$4 = 2^2$ - missing a factor of 2

$6 = 2\times3$ - ✓

$8 = 2^3$ - missing a factor of 2²

$9 = 3^2$ - missing a factor of 3

$10 = 2\times5$ - ✓

$12 = 2^2\times3$ - missing a factor of 2

$14 = 2\times7$ - ✓

$15 = 3\times5$ - ✓

$16 = 2^4$ - missing a factor of 2³

$18 = 2\times3^2$ - missing a factor of 3

$20 = 2^2\times5$ - missing a factor of 2

From the divisors marked "missing", we add the necessary missing factors to the factorization of $M$ , so that

$M = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 2^3 \times 3$

Then the LCM of 1, 2, 3, …, 20 is

$M = 2^4 \times 3^2 \times 5 \times7 \times 11 \times 13 \times17 \times 19$

$\implies \boxed{M = 232,792,560}$

3 0

2 years ago

HELP! 15 POINTS!

Brut [27]

It's pretty easy to go through the choices and none has a[2]=-36 and a[5]=2304. Something's probably wrong with the way the question is typed, but I will answer what's written.

$a_n = a + (n-1) d$

$a_2 = -36 = a + d$

$a_5 = 2304 = a + 4d$

Subtracting,

$2340 = 3d$

$d = 2340/3 = 780$

$a = -36 - d = -36 - 780 = -816$

Answer: a[n] = -816 + 780(n-1) which is none of the above

Check:

$a_2 = -816 + 780= -36 \quad\checkmark$

$a_5 = -816 + 780(4) = 2304 \quad\checkmark$

8 0

3 years ago