All categories

4 years ago

Which figure accurately represents the Pythagorean theorem?

Mathematics

1 answer:

Elena-2011 [213]4 years ago

8 0

Figure B is the answer!

The formula for the Pythagorean theorem is

A^2+B^2=C^2

In figure B the units are squared. Making it your answer.

-Seth

You might be interested in

8-4 skills practice trigonometry

zhannawk [14.2K]

Answer:

what do you need?

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

ANSWER ALL FOUR FOR BRAINLIEST

sergij07 [2.7K]

Answer: (in fractions)

a. 1/4096

b. 1/15625

c. 1/16807

d. 1/432

7 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

If the two factors (below, not the picture tho) 6 and 10,000,000 make 60,000,000 what is the second factor is the equation?

Pachacha [2.7K]

Answer:

It's 6 to the power of 7

Step-by-step explanation:

60,000,000

There are 7 zeros

8 0

3 years ago

How do you substitute x+7y=0 into 2x-8y=22

Grace [21]

First solve for x in the first equation

x + 7y = 0
x = -7y

since x = -7y we can substitute -7y into the other equation wherever there is an x

2x - 8y = 22
2(-7y) - 8y = 22
-14y - 8y = 22
-22y = 22
y = -1

Now let's substitute -1 back into either of the original equations so we can solve for x

x + 7y = 0
x + 7(-1) = 0
x - 7 = 0
x = 7

So your solution set is

(7, -1)

5 0

4 years ago