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3 years ago

liam deposits $2400 into an account that earns 8.3% interest compounded quarterly. how much money will liam have in 5 years.

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

4 0

Answer:

Total = 2,400 * (1 + (.083/4))^4*5

Total = 2,400 * (1.02075)^20

Total = 2,400 * 1.5079528829

Total = 3,619.09

Step-by-step explanation:

nadya68 [22]3 years ago

4 0

Answer:

$3,619.09

Step-by-step explanation:

liam deposits $2400 into an account that earns 8.3% interest compounded quarterly.

Compound interest formula is

$A= P(1+ \frac{r}{n} )^{n*t}$

where A is the final amount

P is the initial amount deposited

r is the rate of interest

n is the compounding period

t is the time in years

P= 2400, r= 8.3%= 0.083, t=5, n=4 (quarterly)

Plug in all the values

$A= 2400(1+ \frac{0.083}{4} )^{4*5}$

A= 3619.09

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What are the domain and range of the function f(x) = 3x + 5

fredd [130]

Step-by-step explanation:

Here,let x = 1,2,3 and 4

Then,

Domain =(1,2,3,4)

Again,

f(1)= 3*1 + 5

=3 + 5

f(2)=3*2 + 5

=6 + 5

=11

f(3)=3*3 + 5

=9 + 5

=14

f(4)=3*4 + 5

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Therefore, Range=(8,11,14,17)

7 0

1 year ago

Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.