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3 years ago

Which statement best explains ?

Mathematics

1 answer:

Tems11 [23]3 years ago

5 0

A is the correct answer. Hope this helps. I need brainliest so if you can I really appreciate it!

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Plz help find the answer plz

Natasha_Volkova [10]

Answer:

$7 {}^{3 + x}$

B. 343 • 7^x

Step-by-step explanation:

if the base of two exponential number is same we add their powers up (only when multiplying)

7^3+7^x ➡7^3+x

the expansion of 7^3 = 343 so 343•7^x is also equivalent to the given expression.

8 0

3 years ago

Read 2 more answers

The amount of increase from the cost to the selling price is called the?

Fynjy0 [20]

Answer:

profit

hope it is helpful to you

5 0

3 years ago

Read 2 more answers

The product of 25 and the number of apples

GREYUIT [131]

"Product" means multiplication
Number of apples we will use the variable "a"

ANSWER: 25a

6 0

3 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

The unit of measure of the of the conversion factor should be the unit of

Usimov [2.4K]

What’s the question?

7 0

4 years ago