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3 years ago

Find two consecutive odd integers whose product is 323

1 answer:

4 0

Use a let statement
first
let x and x + 2 be the number so you write it like this
<u>let x = the first consecutive integer
</u><u>let x + 2 = the second consecutive integer
</u>
second
x(x+2)=323
x^2 + 2x = 323
-323 -323
x^2 + 2x -323 = 0

third
try to factor -323 so it is 19 and -17
(x + 19) (x - 17) = 0
x = 19
x = -17

hope this help

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Need help asap important! thanks so much!

a_sh-v [17]

Answer:

You would have $343.37 at the end of the 2 years.

Step-by-step explanation:

Interest earned is like bonus money the bank pays you just for keeping money

$\mathrm{Compund\:Interest\:Formula}:\quad A = P { \left( 1+ \dfrac{ r }{ n } \right) }^{ nt }$

P: the starting balance of the account (also called initial deposit, or principal)

A: the new balance in the account after N years.

t: the number of years or time

r: the interest rate, (in decimal form)

n: the number of times the interest is compounded each year.

Annually = each year = 1

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Substitute the numbers into the "Compound Interest Formula".

$A = 300 { \left( 1+ \dfrac{ 0.07 }{ 1 } \right) }^{ 1 \times 2 }$

$\mathrm{Anything\:divided\:by\:1\:gives\:itself}$

$A = 300 { \left( 1+0.07 \right) }^{ 1 \times 2 }$

$\mathrm{Add\:1\:and\:0.07\:to\:get\:1.07}$

$A = 300 \times { 1.07 }^{ 1 \times 2 }$

$\mathrm{Multiply\:1\:and\:2\:to\:get\:2}$

$A = 300 \times { 1.07 }^{ 2 }$

$\mathrm{Calculate\:1.07\:to\:the\:power\:of\:2\:and\:get\:1.1449}$

$A = 300 \times 1.1449$

$\mathrm{Multiply\:300\:and\:1.1449\:to\:get\:343.47}$

$A = 343.47$

So you would have $343.37 at the end of the 2 years.

Look at the chart

$\mathrm{Year}\quad \mathrm{Year\: Interest}\quad \mathrm{Total\:Interest}\quad \mathrm{Balance}\\\quad1\quad\quad\quad $21.00\quad\quad\quad\quad $21.00\quad\quad\quad\quad $321.00\\\quad2\quad\quad\quad $22.47\quad\quad\quad\quad $43.47\quad\quad\quad\quad $343.47$

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