Let the number of text messages be represented by X.
Then the total cost of David's cell phone adds up to
Since the total cost of David's phone equals $91.40, we get the following equation.
Subtract 50 on both sides
Divide both sides by 0.45
Therefore David has send 92 text messages.
Answer:
142
Step-by-step explanation:
60+70=130
8+4=12
130+12=142
keeping in mind that
x = percent rate for the 17000 investment.
y = percent rate for the 11000 investment.
so the amount for the 17000 interest will just be (x/100) * 17000, or namely 170x.
and the amount of interest earned for the 11000 is (y/100) * 11000, or just 110y.
now, regardless of what "x" and "y" are, we know that the interest from the 17000 is higher by 308 bucks, therefore 170x = 110y + 308.
we also know that the rate of <u>x</u> is higher as well than <u>y</u> by 0.4%, so then x = y + 0.4.
![\bf \begin{cases} 170x=110y+308\\ \boxed{x}= y +0.4\\[-0.5em] \hrulefill\\ 170\left( \boxed{y+0.4} \right)=110y+308 \end{cases} \\\\\\ 170y+68=110y+308\implies 60y=240\implies y=\cfrac{240}{60}\implies \blacktriangleright y=\stackrel{\%}{4} \blacktriangleleft \\\\\\ x=y+0.4\implies \blacktriangleright x=\stackrel{\%}{4.4} \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20170x%3D110y%2B308%5C%5C%20%5Cboxed%7Bx%7D%3D%20y%20%2B0.4%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20170%5Cleft%28%20%5Cboxed%7By%2B0.4%7D%20%5Cright%29%3D110y%2B308%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20170y%2B68%3D110y%2B308%5Cimplies%2060y%3D240%5Cimplies%20y%3D%5Ccfrac%7B240%7D%7B60%7D%5Cimplies%20%5Cblacktriangleright%20y%3D%5Cstackrel%7B%5C%25%7D%7B4%7D%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20x%3Dy%2B0.4%5Cimplies%20%5Cblacktriangleright%20x%3D%5Cstackrel%7B%5C%25%7D%7B4.4%7D%20%5Cblacktriangleleft)
Answer:
rule: f(n) = n³
missing numbers: 125, 216, 343, 512, 729
Step-by-step explanation:
Your familiarity with the cubes of small integers helps you recognize each of these numbers is a cube. Their sequence is the sequence of cubes of increasing natural numbers.
1 = 1·1·1 = 1³
8 = 2·2·2 = 2³
27 = 3·3·3 = 3³
64 = 4·4·4 = 4³
__
The rule is ...
f(n) = n³
The cubes of 5 through 9 will complete the set of numbers ...
1, 8, 27, 64, 125, 216, 343, 512, 729
