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GarryVolchara [31]

3 years ago

7

I WILL GIVE BRAINLIST IF YOU HELP

1 answer:

Dvinal [7]3 years ago

6 0

Answer:

I would say the last option

Step-by-step explanation:

If you compare it, the 8 is around half of the radius, so 8 doubled would be 16, and the 28 in half would be 14, and this number needs to be a bit bigger since that was placed higher than the diameter.

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You buy 6 movie tickets and a $7.50 bucket of popcorn. If you spend a total of $115.50 at the movies, how much was each ticket?

Norma-Jean [14]

Answer:

Mi novia me terminó por mi hermano

Step-by-step explanation:

8 0

3 years ago

B) Alex invests $2000 in an account that has a 6% annual rate of growth. To the nearest year, when

wlad13 [49]

Answer:

part 1) 10 years

part 2) 10 years

Step-by-step explanation:

<u><em>The correct question is:</em></u>

Part 1) Alex invests $2000 in an account that has a 6% annual rate of growth compounded annually. To the nearest year, when will the investment be worth $3600?

Part 2) Alex invests $2000 in an account that has a 6% annual rate of growth compounded continuously. To the nearest year, when will the investment be worth $3600?

Part 1) 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

in this problem we have

$P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06\\n=1$

substitute in the formula above

$3,600=2,000(1+\frac{0.06}{1})^{t}$

$1.8=(1.06)^{t}$

Apply log both sides

$log(1.8)=log[(1.06)^{t}]$

Applying property of exponents

$log(1.8)=(t)log(1.06)$

$t=log(1.8)/log(1.06)$

$t=10.09\ years$

Round to the nearest year

$t=10\ years$

Part 2) 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

$P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06$

substitute in the formula above

$3,600=2,000(e)^{0.06t}$

$1.8=[e]^{0.06t}$

Apply ln both sides

$ln(1.8)=ln[e]^{0.06t}$

$ln(1.8)=(0.06t)ln[e]$

$t=ln(1.8)/0.06$

$t=9.80\ years$

Round to the nearest year

$t=10\ years$

6 0

3 years ago

How many solutions does the system have?

klasskru [66]

Answer:

wait ok

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

B) Given y= 5x – 4, enter the value of x when y = –39

Serggg [28]

Answer:

try using 7 for your anwser i hope this helps

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is 10 * 10 *100 because it is hard?

Romashka-Z-Leto [24]

10x10x100=
10000
Would be the answer hope this helped.

7 0

3 years ago