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Bezzdna [24]
3 years ago
11

work out three numbers that are in both sequences and between 20 and 40 write the numbers on a single line

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

The nth term of a sequence is 2n + 1

The nth term of a different sequence is 3n  1

Work out the three numbers that are

in both sequences

and

between 20 and 40

Now the three numbers I managed to get are 25, 31 and 37 (as I assumed it meant the output of both of these sequences needs to be the same as well as being in between 20 and 40). For example 2x18+1= 37 as does 3x12+1= 37

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- 9y2 (2y2 + 3y - 9)
Harrizon [31]

Answer:

=−18y4−27y3+81y2

Step-by-step explanation:

5 0
3 years ago
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What is the slope and y intercept of this line? y = -4x + 7
amid [387]
Slope is -4 and y intercept is 7. Whatever the number in front of the X is, that will always be your slope. Anything added or subtracted after the x, that’s your y intercept! Hope I helped.
4 0
2 years ago
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A florist has to choose four different types of flowers to include in a bouquet. In how many ways can the florist do this, if th
Nat2105 [25]

Answer:

Step-by-step explanation:

Given:

Type of Flowers = 5

To choose = 4

Required

Number of ways 4 can be chosen

The first flower can be chosen in 5 ways

The second flower can be chosen in 4 ways

The third flower can be chosen in 3 ways

The fourth flower can be chosen in 2 ways

Total Number of Selection = 5 * 4 * 3 * 2

Total Number of Selection = 120 ways;

Alternatively, this can be solved using concept of Permutation;

Given that 4 flowers to be chosen from 5,

then n = 5 and r = 4

Such that

nPr = \frac{n!}{(n - r)!}

Substitute 5 for n and 4 for r

5P4 = \frac{5!}{(5 - 4)!}

5P4 = \frac{5!}{1!}

5P4 = \frac{5*4*3*2*1}{1}

5P4 = \frac{120}{1}

5P4 = 120

Hence, the number of ways the florist can chose 4 flowers from 5 is 120 ways

4 0
3 years ago
Lcm of (x-2)(x+3) and 10(x+3)^2
Arte-miy333 [17]
Least common multiple: factor them, then see what they have in common and what is leftover and multiply those expressions:

(x - 2)(x + 3)          10(x + 3)(x + 3)
Common: (x + 3)
Leftover: (x - 2), (10), (x + 3)
Common · Leftover is: (x + 3) · (x - 2) · (10) · (x + 3) = 10(x - 2)(x + 3)²

Answer: LCM is 10(x - 2)(x + 3)²


8 0
3 years ago
Is My answer correct?
Licemer1 [7]
Yes,  you are correct!
6 0
3 years ago
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