[tex](1 imes rac{2}{7} - rac{1}{3} ) imes 1 imes rac{3}{4} [/tex]

Mathematics, 14.07.2019 12:30 bryamp4268

$(1 \times \frac{2}{7} - \frac{1}{3} ) \times 1 \times \frac{3}{4}$

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Answer from: connorgking

I assume you want to evaluate the expression. So, let's use the priority rules properly: first of all, parenthesis. So, for now, we concentrate on

$1 \times \cfrac{2}{7}-\cfrac{1}{3}$

We again use the priority rules to remember that we need to perform multiplications first, and then additions/subtractions. So, first of all we have to evaluate

$1 \times \cfrac{2}{7} = \cfrac{2}{7}$

Since every number multiplied by 1 remains unchanged. The expression in parenthesis is simplified to

$\cfrac{2}{7}-\cfrac{1}{3}$

Transform the denominators of both fractions to 21 in order to subtract them:

$\cfrac{2}{7}\cdot\cfrac{3}{3}-\cfrac{1}{3}\cdot\cfrac{7}{7} = \cfrac{6}{21}-\cfrac{7}{21} = -\cfrac{1}{21}$ .

Now that we simplified the expression in the parenthesis, we can focus on the whole expression again, which has become

$-\cfrac{1}{21} \times 1 \times \cfrac{3}{4}$

We're left with multiplications only, so we can perform them in any order, since the multiplication is commutative. As before, we can ignore the multiplication by 1, since it doesn't change the value of the expression:

$-\cfrac{1}{21} \times \cfrac{3}{4}$

Finally, to multiply two fractions, you simply multiply numerators and denominators among them:

$-\cfrac{1}{21} \times \cfrac{3}{4} = -\cfrac{1\cdot3}{21\cdot4} = -\cfrac{3}{84} = -\cfrac{1}{28}$

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

Answer from: Quest

answer: your answer is c

step-by-step explanation:

Asap a b c or d **if you know it answer, if not don’t waist my time**

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[tex](1 \times \frac{2}{7} - \frac{1}{3} ) \times 1 \times \frac{3}{4} [/tex]...

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