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3 years ago

1.075(5.25)x - 5.25(0.25)(x-3)

Mathematics

1 answer:

Vitek1552 [10]3 years ago

5 0

For this case we must simplify the following expression:

$1,075 (5.25) x-5.25 (0.25) (x-3) =$

So, solving the first parentheses we have:

$5.64375x-1.3125 (x-3) =$

We have to:

$- * + = -\\- * - = +\\5.64375x-1.3125x + 1.3125 (3) =\\5.64375x-1.3125x + 3.9375 =$

We add similar terms taking into account that different signs are subtracted and the sign of the major is placed:

$4.33125x + 3.9375$

Finally, the simplified expression is: $4.33125x + 3.9375$

ANswer:

$4.33125x + 3.9375$

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3 years ago

Let an be the sum of the first n positive odd integers.

sesenic [268]

Answer:

A) The first 4 terms of the sequences are: $a_{1} =16$ , $a_{2} =24$ , $a_{3} =32$ and $a_{4} =40$ .

B) An explicit formula for this sequence can be written as: $a_{n} =8*(n+1)$

C) A recursive formula for this sequence can be written as:

$\left \{ {{a_{1} =16} \atop {a_{n} =a_{n-1}+8}} \right.$

Step-by-step explanation:

A) You can find the firs terms of this sequence simply selecting an odd integer and summing the consecutive 3 ones:

$a_{n} = Odd_{n}+Odd_{n+1}+Odd_{n+2}+Odd_{n+3}$ (a.1)

$a_{1}=1+3+5+7=16$

$a_{2}=3+5+7+9=24$

$a_{3}=5+7+9+11=32$

$a_{4}=7+9+11+13=40$

B) Observe the sequence of odd numbers 1, 3, 5, 7, 9, 11, 13(...).

You can express this sequence as:

$Odd_{n}=(2*n-1)$ (b.1)

If you merge the expression b.1 in a.1, you obtain the explicit formula of the sequence:

$a_{n} = Odd_{n}+Odd_{n+1}+Odd_{n+2}+Odd_{n+3}$ (a.1)

$a_{n} = (2*n-1)+((2*(n+1)-1))+((2*(n+2)-1))+((2*(n+3)-1))$ (b.2)

$a_{n} = 8*n+8$ (b.3)

$a_{n} =8*(n+1)$ (b.s)

C) The recursive formula has to be written considering an initial term and an N term linked with the previous term. You can see an addition of 8 between a term and the next one. So you can express each term as an addition of 8 with the previous one. Therefore, if the first term is 16:

$\left \{ {{a_{1} =16} \atop {a_{n} =a_{n-1}+8}} \right.$ (c.s)

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3 years ago

When adding like fractions,why is the numerator the only part of the fraction that changes?

dolphi86 [110]

Because youre adding parts of the fraction and not changing the amount of parts the fraction is divided in to.

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Read 2 more answers

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lilavasa [31]

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