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3 years ago

Suppose given parallelograms EFGH ∼ IJKL, such that m∠E=70°. Find m∠J and m∠L.

Mathematics

1 answer:

jarptica [38.1K]3 years ago

8 0

if both parallelograms are similar, that means all their angles are the same, one might be larger or smaller, but the angles are all the same for both.

in a parallelogra, opposite angles are equal, and their counterpart angle, are their supplementary sibling, namely they both siblings add to 180°.

check the picture below

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An employee joined a company in 2009 with a starting salary of $50,000. Every year this employee receives a raise of $1000 plus

stepladder [879]

Answer:

(a) The required recurrence relation for the salary of the employee of n years after 2009 is $a_n=1.05a_{n-1}+1000$ .

(b)The salary of the employee will be $83421.88 in 2017.

(c) $\therefore a_n=70,000 . \ 1.05^n-20,000$

Step-by-step explanation:

Summation of a G.P series

$\sum_{i=0}^n r^i= \frac{r^{n+1}-1}{r-1}$

(a)

Every year the salary is increasing 5% of the salary of the previous year plus $1000.

Let $a_n$ represents the salary of the employee of n years after 2009.

Then $a_{n-1}$ represents the salary of the employee of (n-1) years after 2009.

Then $a_n= a_{n-1}+5\%.a_{n-1}+1000$

$=a_{n-1}+0.05a_{n-1}+1000$

$=(1+0.05)a_{n-1}+1000$

$=1.05a_{n-1}+1000$

The required recurrence relation for the salary of the employee of n years after 2009 is $a_n=1.05a_{n-1}+1000$ .

(b)

Given, $a_0=\$50,000$

$a_n=1.05a_{n-1}+1000$

Since 2017 is 8 years after 2009.

So, n=8.

∴ $a_8$

$=1.05 a_7+1000$

$=1.05(1.05a_6+1000)+1000$

$=1.05^2a_6+1.05\times 1000+1000$

$=1.05^2(1.05a_5+1000)+1.05\times 1000+1000$

$=1.05^3a_5+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^3(1.05a_4+1000)+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^4a_4+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^4(1.05a_3+1000)+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^5a_3+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^5(1.05a_2+1000)+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^6a_2+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^6(1.05a_1+1000)+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^7a_1+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^7(1.05a_0+1000)+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^8a_0+1.05^7\times1000+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^8a_0+(1.05^7+1.05^6+1.05^5+1.05^4+1.05^3+1.05^2+1.05+1)1000$

$=1.05^8 \times 50,000+\frac{1.05^8-1}{1.05-1}\times 1000$

$=1.05^8\times 50,000+20,000(1.58^8-1)$

$=70,000\times 1.05^8-20,000$

≈$83421.88

The salary of the employee will be $83421.88 in 2017.

(c)

Given, $a_0=\$50,000$

$a_n=1.05a_{n-1}+1000$

We successively apply the recurrence relation

$a_n=1.05a_{n-1}+1000$

$=1.05^1a_{n-1}+1.05^0.1000$

$=1.05^1(1.05a_{n-2}+1000)+1.05^0.1000$

$=1.05^2a_{n-2}+1.05^1.1000+1.05^0.1000$

$=1.05^2(1.05a_{n-3}+1000)+(1.05^1.1000+1.05^0.1000)$

$=1.05^3a_{n-3}+(1.05^2.1000+1.05^1.1000+1.05^0.1000)$

...............................

.................................

$=1.05^na_{n-n}+\sum_{i=0}^{n-1}1.05^i.1000$

$=1.05^na_0+1000\sum_{i=0}^{n-1}1.05^i$

$=1.05^n.50,000+1000.\frac{1.05^n-1}{1.05-1}$

$=1.05^n.50,000+20,000.(1.05^n-1)$

$=(50,000+20,000)1.05^n-20,000$

$=70,000 . \ 1.05^n-20,000$

$\therefore a_n=70,000 . \ 1.05^n-20,000$

6 0

3 years ago