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1 year ago

Natalie invested $40,000 in an account paying an interest rate of 8 % compounded

Mathematics

1 answer:

NARA [144]1 year ago

5 0

Answer:

8000

Step-by-step explanation:

Natalie

Divide 40k by 2 to get 20k and divide 20k by 5 to get 4k we are at 10%

4k / 5 = 800 that means that every 2% is 800 now we multiply by 4x

To get 8% and we get 3200 and multiply by 20 representing the yrs

3200 x 20 = 64k + 40+ = 104k

Mackenzie

40k divide by 2 to get 50% 40k / 2 = 20k and now we divide by 5 to get 10% 20k / 5 = 4k now we divide by 10 to get the amount of each percents worth 4k / 10 = 400 we multiply by 9 to get the compounded interest

400 x 9 = 3600 multiply by 20 to get the compounded daily investment after 20yrs 3600 x 20 = 72k + 40k = 112k

112k - 104k = 8k

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What is the area of this parallelogram

Umnica [9.8K]

Answer:

112 m^2

Step-by-step explanation:

A = b * h

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7 0

3 years ago

Read 2 more answers

How do you do this? Please help

PtichkaEL [24]

Answer:

$PO=86.2\\ OM=23.8\\$

Step-by-step explanation:

To preface, your figure is going to be a line segment, with $O$ as your midpoint, in between points $P$ & $M.$

With that being said:

$PO+OM=PM$

Identify your values:

$PO=7y+12\\OM=3y-8\\PM=110$

Substitute the values into the first equation:

$7y+12+3y-8=110$

Combine like terms:

$10y+4=110$

Subtract $4$ from both sides of the equation:

$10y=106$

Divide by the coefficient of $y$ , which is $10$ :

$y=10.6$

Substitute $10.6$ for $y$ in segments $PO$ & $OM$ :

$PO=7(10.6)+12$

$OM=3(10.6)-8$

$PO=74.2+12$

$OM=31.8-8$

Solve:

$PO=86.2$

$OM=23.8$

Check your answers by substituting:

$PO+OM=PM$

$86.2+23.8=110$

8 0

3 years ago

Anyone know how to solve this??

Tom [10]

$Given : \frac{(x^\frac{2}{5})^9\;.\;(x^\frac{-4}{15})}{x^\frac{1}{3}}$

$\implies {(x^\frac{18}{5})\;.\; (x^\frac{-4}{15})}\;.\;(x^\frac{-1}{3})$

$\implies x^(^\frac{18}{5} ^-^\frac{4}{15}^-^\frac{1}{3}^)$

$\implies x^(^(^\frac{54 - 4}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{50}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{10}{3}^)^-^\frac{1}{3}^) = x^(^\frac{9}{3}^) = x^3$