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3 years ago

I need help on 10 13 16

Mathematics

1 answer:

Zepler [3.9K]3 years ago

3 0

$10.\\1ft=0.3048m\to1m=\dfrac{10000}{3048}ft\approx3.281ft\\\\therefore\ 14m\approx14\cdot3.28ft1=45.93ft$

$13.\\1mi=1.609km\to1km=\dfrac{1000}{1609}mi\approx0.621mi\\\\therefore\ 0.3km\approx0.3\cdot0.621mi=0.186mi$

$1in=2.54cm\to1cm=\dfrac{100}{254}in\approx0.394in\\\\therefore\ 15cm\approx15\cdot0.394in=5.91in$

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Step-by-step explanation:

Add common terms x and y together

10x^3+3y^3+11x^2-1

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4 years ago

Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily,

Rudik [331]

Answer:

Part 1) 8.17 years

Part 2) 4.98 years

Part 3) 4.95 years

Part 4) 4.95 years

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part 1) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded annually

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=1$

substitute in the formula above

$2p=p(1+\frac{0.14}{1})^{t}$

$2=(1.14)^{t}$

Apply log both sides

$log(2)=log[(1.14)^{t}]$

$log(2)=(t)log(1.14)$

$t=log(2)/log(1.14)$

$t=8.17\ years$

Part 2) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded monthly

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=12$

substitute in the formula above

$2p=p(1+\frac{0.14}{12})^{12t}$

$2=(\frac{12.14}{12})^{12t}$

Apply log both sides

$log(2)=log[(\frac{12.14}{12})^{12t}]$

$log(2)=(12t)log(\frac{12.14}{12})$

$t=log(2)/12log(\frac{12.14}{12})$

$t=4.98\ years$

Part 3) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded daily

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=365$

substitute in the formula above

$2p=p(1+\frac{0.14}{365})^{365t}$

$2=(\frac{365.14}{365})^{365t}$

Apply log both sides

$log(2)=log[(\frac{365.14}{365})^{365t}]$

$log(2)=(365t)log(\frac{365.14}{365})$

$t=log(2)/365log(\frac{365.14}{365})$

$t=4.95\ years$

Part 4) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% continuously

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14$

substitute in the formula above

$2p=p(e)^{0.14t}$

Simplify

$2=(e)^{0.14t}$

Apply ln both sides

$ln(2)=ln[(e)^{0.14t}]$

$ln(2)=(0.14t)ln(e)$

Remember that ln(e)=1

$ln(2)=(0.14t)$

$t=ln(2)/(0.14)$

$t=4.95\ years$

4 0

3 years ago

Mr Nersin likes to ride his bike. He rode for 1/2 hour. After a short rest, he rode 1/10 hour more. He traveled 4 miles in all.

vodomira [7]

The distance traveled by Mr.Nersin = 4 miles
The time for which Mr.Nersin rode first = 1/2 hour
The time for which Mr.Nersin rode after a short rest = 1/10 hour
Then
The total time for which Mr.Nersin rode = (1/2) + (1/10) hour
= (5 + 1)/10 hour
= 6/10 hour
= 3/5 hour
So
Average speed of Mr. Nersin = Distance traveled / Time taken
= 4/(3/5) miles/hour
= (4 * 5)/3 miles/hour
= 20/3 miles/hour
= 6 2/3 miles/hour
= 6.67 miles per hour
So the average speed at which Mr.Nersin traveled is 6 2/3 or 6.67 miles per hour.

6 0

3 years ago

Read 2 more answers