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nasty-shy [4]
3 years ago
9

How much would $600 invested at 8% interest compounded continuously worth after 3 years?

Mathematics
1 answer:
densk [106]3 years ago
6 0

Answer:

a \:  =  \: pe ^{rt}

a = 600e^{(.08)(3)}

a = 762.75 \: dollars

Since the question stated that the interest is compounded continuously, we know that Euler number is required

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when x equals one number, if it was no solution then x=y

Step-by-step explanation:

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What is the solution to the system of equations Use the substitution method to solve the system of equations. Show your work.
marishachu [46]

Answer:

Step-by-step explanation:

Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x        Equation 1

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24        Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

Here's an idea! Equation 11 tells us that \goldD{2x}2x and \goldD yy are equal. So let's plug in \goldD{2x}2x for \goldD yy in Equation 22 to get rid of the yy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}  

x+y

x+2x

​    

=24

=24

​    

Equation 2

Substitute 2x for y

​  

Brilliant! Now we have an equation with just the xx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}  

y

y

y

​    

=2x

=2(8)

=16

​    

Equation 1

Substitute 8 for x

​  

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}  

y

16

16

​    

=2x

=

?

2(8)

=16

​    

Plug in x = 8 and y = 16

Yes!

​  

Let's check the second equation:

\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}  

x+y

8+16

24

​    

=24

=

?

24

=24

​    

Plug in x = 8 and y = 16

Yes!

​  

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

Your turn to solve a system of equations using substitution.

Use substitution to solve the following system of equations.

4x + y = 284x+y=28

y = 3xy=3x

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How many oranges do you need for 1 cup
mel-nik [20]
You will need 2 oranges for 1 cup
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Data are collected on the 35 students in a college history course. Which of the following is not a variable for the data set?
Nesterboy [21]

Answer:

(E) Number of students in the data set

Explanation:

A variable, in research and data collection, refers to something that is being measured and can have changing values.  

The example above shows that student birth month is a variable since we can have a range of options from January to December. Political affiliation is also a variable since the students can state whether they follow certain political parties or even whether they do not. Student age is a variable too, with answers from a range of numbers such as 20-25, since they are college students. Student address is a variable as well since students will have varying answers, be it those who live in the same address or different ones.  

However, number of students in the data set is not a variable – it is instead the number of research participants. It can be a population or a sample, depending on what the research is about.  

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3 years ago
Discriminat step by step of x^2+4x+5=0
alisha [4.7K]

Answer:

-4

Step-by-step explanation:

x^2+4x+5=0

a=1,b=4^1 C=5

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