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3 years ago

The intersection of the perpendicular rays below has formed which of the following?

Mathematics

1 answer:

dlinn [17]3 years ago

4 0

B. 4 right angles

perpendicular lines always create 90° angles

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In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose

viktelen [127]

Answer:

3 students

Step-by-step explanation:

If everyone in the class has either a pierced nose or ear, we just simply have to add up the total number of hands raised and minus the number of students in the class.

35+8=43

43-40=3

3 students have both a pierced nose and pierced ear.

6 0

2 years ago

You find an interest rate of 10% compounded quarterly. Calculate how much more money you would have in your pocket if you had us

Elena-2011 [213]

Answer:

see the explanation

Step-by-step explanation:

we know that

step 1

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

$r=10\%=10/100=0.10\\n=4$

substitute in the formula above

$A=P(1+\frac{0.10}{4})^{4t}$

$A=P(1.025)^{4t}$

Applying property of exponents

$A=P[(1.025)^{4}]^{t}$

$A=P(1.1038)^{t}$

step 2

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

$r=10\%=10/100=0.10$

substitute in the formula above

$A=P(e)^{0.10t}$

Applying property of exponents

$A=P[(e)^{0.10}]^{t}$

$A=P(1.1052)^{t}$

step 3

Compare the final amount

$P(1.1052)^{t} > P(1.1038)^{t}$

therefore

Find the difference

$P(1.1052)^{t} - P(1.1038)^{t}$ ----> Additional amount of money you would have in your pocket if you had used a continuously compounded account with the same interest rate and the same principal.