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3 years ago

Help me out here please

Mathematics

1 answer:

Maslowich3 years ago

7 0

4 5 8 11

5+8/2 = 6.5

Hope this helps!

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Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

<em>Solving the expression</em>

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

<em>Solving the expression</em>

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

<em>Solving the expression</em>

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

<em>Solving the expression</em>

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

#learnwithBrainly

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3 years ago

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PLS HELP

Anika [276]

The chairs in the auditorium illustrates an arithmetic sequence

The recursive rule is: an = an-1 + 8; a1 = 26
The explicit rule is: an = 18 + 8n
There are 74 seats in the 7th row
840 chairs can fit the auditorium

<h3>The recursive rule</h3>

From the question, we have the following sequence:

Seats: 18, 26, 34

Rewrite as:

26 = 18 + 8

34 = 26 + 8

The above means that:

The number of seats on a row is 8 added to the the number of seats on the previous row.

Hence, the recursive rule is:

an = an-1 + 8; a1 = 26

<h3>The explicit rule</h3>

In (a), we have:

a1 = 26 and d = 8

The explicit rule is calculated using:

an = a1 + (n -1) * d

This gives

an = 26 + (n - 1) * 8

Expand

an = 26 - 8 + 8n

Evaluate the difference

an = 18 + 8n

Hence, the explicit rule is:

an = 18 + 8n

<h3>The number of seats in the row 7</h3>

This means that n = 7.

So, we have:

a7 = 18 + 8 * 7 = 74

Hence, there are 74 seats in the 7th row

<h3>The sigma notation</h3>

We have the maximum number of rows to be 12.

So, the sigma notation would be:

$\sum\limits^{12}_{n=1} {18 + 8n}$

The total number of seats is:

Sn = n/2(2a + (n -1) * d)

This gives

S12 = 12/2(2 * 26 + (12 -1) * 8)

S12 = 840

Hence, 840 chairs can fit the auditorium

Read more about arithmetic sequence at:

brainly.com/question/6561461

7 0

3 years ago