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makkiz [27]
1 year ago
5

CAN SOMEONE PLEASE HELP ME PLEASE!!!!!

Mathematics
1 answer:
OverLord2011 [107]1 year ago
7 0

Answer:

x4+7x3-2x2-9x-3 remainder -10

or we could write it as

x4+7x3-2x2-9x-3 - 10/(x+8)

Step-by-step explanation:

x+8 ) x5+15x4+54x3−25x2−75x−34 (x4+7x3-2x2-9x-3  <--- Quotient.

        x5+ 8x4

                7x4+54x3

                7x4+56x3

                         -2x3-25x2

                         -2x3-16x2

                                    -9x2-75x

                                    -9x2-72x

                                               -3x-34

                                               -3x-24

                                                     -10  <--- Remainder.

 

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Its either -1,2 or 1,2
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3 years ago
The length of a rectangle is 3cm longer than its width.
Contact [7]

Answer:

width=8cm

lenth=11

Step-by-step explanation:

8 0
3 years ago
Pythagorean Theorem a=11 b=60 c=?
zhuklara [117]

Answer:

61

Step-by-step explanation:

The Pythagorean Theorem is a fundamental in geometry that involves looking at two sides of a right triangle and finding out the length of it. Let's use the formula with the corresponding numbers.

The formula for this question would be c=\sqrt{a^2 + b^2}

Since we know that a = 11 and b = 60, we can do the following and substitute :

c = \sqrt{11^2 + 60^2}

11 squared is 121

60 squared is 3600

Now we need to add the two number together :

121 + 3600

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Since this number is under the square root, we have to square root it. :

\sqrt{3721}

61

Therefore c is equal to 61.

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3 years ago
Three people who work full-time are to work together on a project, but their total time on the project is to be equivalent to th
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3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
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