The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let \displaystyle PP be the student population and \displaystyle nn be the number of years after 2013. Using the explicit formula for a geometric sequence we get
{P}_{n} =284\cdot {1.04}^{n}P
n
=284⋅1.04
n
We can find the number of years since 2013 by subtracting.
\displaystyle 2020 - 2013=72020−2013=7
We are looking for the population after 7 years. We can substitute 7 for \displaystyle nn to estimate the population in 2020.
\displaystyle {P}_{7}=284\cdot {1.04}^{7}\approx 374P
7
=284⋅1.04
7
≈374
The student population will be about 374 in 2020.
Answer:
The answer is x = -4
Step-by-step explanation:
~Hoped this helped~
Treat the matrices on the right side of each equation like you would a constant.
Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.
Then you can eliminate <em>Y</em> by taking the sum
4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>
==> <em>X</em> = (4<em>A</em> + <em>B</em>)/11
Similarly, you can eliminate <em>X</em> by using
-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>
==> <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11
It follows that
Similarly, you would find
You can solve the second system in the same fashion. You would end up with
Answer:
where are the question??????
Answer:
Step-by-step explanation:
number of
pennies=8 % of 50=8/100×50=4
dimes=40 % of 50=40/100×50=20
nickels=4+3=7
quarters=50-(4+20+7)=50-31=19
total money=4+20×10+7×5+19×25
=4+200+35+475
=714 pennies
=7.14 $
