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3 years ago

PPPLLLZZZ HELP I'LL GIVE BRAINLIEST

Mathematics

2 answers:

Yuliya22 [10]3 years ago

8 0

Hello,

Original problem) –8.3 + 9.2 – 4.4 + 3.7.

Step 1) −8.3 + 9.2 + 4.4 + 3.7 Additive inverse

Step 2) −8.3 + 4.4 + 9.2 + 3.7 Commutative property

Step 3) −8.3 + (4.4 + 9.2 + 3.7) Associative property

Step 4) −8.3 + 17.3

Allen should have collected negative numbers together and positive numbers together.

Hope this helps!!!! (:

kati45 [8]3 years ago

6 0

Allen's work is not written properly so I have rearranged it as shown below:

Original problem) –8.3 + 9.2 – 4.4 + 3.7.

Step 1) −8.3 + 9.2 + 4.4 + 3.7 Additive inverse

Step 2) −8.3 + 4.4 + 9.2 + 3.7 Commutative property

Step 3) −8.3 + (4.4 + 9.2 + 3.7) Associative property

Step 4) −8.3 + 17.3

We can see that in step 1), Allen changed -4.4 into +4.4 using additive inverse. Notice that we are simplifying not eliminating -4.4 as we do in solving some equation. Hence using additive inverse is the wrong step.

Alen should have collect negative numbers together and positive numbers together.

Add the respective numbers then proceed to get the answer.

–8.3 + 9.2 – 4.4 + 3.7

= –8.3 – 4.4 + 9.2 + 3.7

= -12.7 + 12.9

= 0.2

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How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

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Mashcka [7]

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OOOOOHHH a breakthrough! I was confused when I read this question but now I get it!

Okay when dividing this equation I would do this

5^4

Since 25 is 5 x 5 I would mark it as 5^2, and then I can subtract!

5^4

5^2 =

5^2

If you do it like this then it would be the first choice.

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