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4 years ago

What is the pattern in the values as the exponents increase?

Mathematics

1 answer:

Anastasy [175]4 years ago

4 0

Answer:

Option D is correct.

Step-by-step explanation:

The first value is 1/32

if we multiply 1/32 with 2 i.e 1/32*2 we get 1/16

So, next value is 1/16

if we multiply 1/16 with 2 i.e 1/16 * 2 we get 1/8

So, next value is 1/8

if we multiply 1/8 with 2 i.e 1/8 * 2 we get 1/4

So, next value is 1/4

if we multiply 1/4 with 2 i.e. 1/4 *2 we get 1/2

so, next value is 1/2

if we multiply 1/2 with 2 i.e 1/2 * 2 we get 1

So, Option D multiply the previous value by 2 is correct.

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Maxwell purchased $15,000 worth of 52-week T-Bills for $14,650. What will be the rate of return on his investment?

Ket [755]

Answer:

2.39%

Step-by-step explanation:

Maxwell bought $15,000 worth of 52-week T-bills for $14,650

Therefore the rate of return on investment can be calculated as follows

= 15,000-14,650/14,650× 100

= 350/14,650 × 100

= 0.02389 × 100

= 2.39%

Hence the rate of return on this investment is 2.39%

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3 years ago

Enter the number of complex zeros for the polynomial function in the box.

NikAS [45]

Since the equation has a degree of 5 then, it will definitely have 5 zeroes. Complex zeroes of the polynomial always come in pair such that even if there are five zeroes, the complex zeroes could only be either 2 or 4 or none.

Hence, the answer to this item is none, 2 or 4.

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3 years ago

Read 2 more answers

Solve for k.

EastWind [94]

To do this we need to rearrange the equation.
m = -8j - 9k + 9
Subtract (-8j + 9) from both sides
m + 8j - 9 = -9k
Divide both sides by -9
(m + 8j - 9)/-9 = k

The answer is C

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

A circle centered at the origin has a radius of 5 units. The terminal side of an angle, , intercepts the circle in Quadrant 3 at

rodikova [14]

Answer:

sorry

Step-by-step explanation:

5 0

3 years ago