Exponential functions

Mathematics

2 answers:

Marysya12 [62]2 years ago

6 0

Answer:

2,323.23

Step-by-step explanation:

the interest will be 323.23 +your initial balance is what you will have in 5 years

Damm [24]2 years ago

6 0

In order to figure out the compound interest, you must first identify its equation which is: A = P(1+(r/n))^nt where A is the final amount, P is the initial amount, R is the interest rate, n is the number of times the interest is applied per period, and t is the number of periods. Plugging in your values would make the equation: A = 2000(1+(0.03/12))^(12•5) simplified would be A = 2323.23

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How is |3+10| = |3| + |10| equal

Morgarella [4.7K]

Answer:

First we must figure out what the question means. It is asking how the absolute value of 3+10 is equivalent to the absolute value of 3 plus the absolute value of ten.

Next we should solve the expressions: The absolute value of 3 plus 10 is 13. The absolute value of 3 plus the absolute value of 10 is 13. 13 and 13 are the same number.

Therefore, the expressions are equivalent. Thus, having your answer.

8 0

2 years ago

If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num

weeeeeb [17]

Answer:

Step-by-step explanation:

$x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right$

$\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right$

$\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}$