STruggling. pls help

Can someone please answer these two questions? Isn't the first one's answer option A? I don't understand the second one. Please

eimsori [14]

Answer:

no 2 answer is A(8)

and no 3 answer is 5m-1

i have read this

3 0

3 years ago

Ajaaaaaaaaaaaaaaaaaaa<br> 28181x6

lilavasa [31]

Answer:

169086

Step-by-step explanation:

28181 x 6 = 169086

Use a calculator?

8 0

3 years ago

Which equation is true? Explain your answer.

Liula [17]

Answer:

Step-by-step explanation:

1. Find if all of the fractions are equal to the other fraction next to them.

First, let's go with the denominator of all.

6/8 = 4/3

First of all, we can easily say that this is not right, since the fact that, 6 is less than 8, and 4 is greater than 3. That's just signaling that this can't be the answer.

3/5 = 8/10

10/5=2

8/3=??? (doesn't go into 8)

So, this one is also wrong

12/16 = 36/48

48/16=3

36/12=3

<h2>So hence, the last one or 12/16=36/48 is RIGHT.</h2>

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Thanks!

Answered by: FieryAnswererGT

#learnwithbrainly

Want to thank me? MARK ME BRAINLIEST! I'll get happy, you will get happy! EQUAL!

3 0

2 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

Recycled cd’s incorporated, offers a choice of 5 used CD for $24, with each additional CD costing $5. Write a cost function for

DiKsa [7]

If $24 represents the cost of the 5 used CDs, than your formula would be "24+5x" ,x being the amount of extra CDs you purchase.

24+5(6-5)

24+5(1)

24+5

=29

6 CDs would cost $29

I hope thats right, goodluck^^

5 0

3 years ago