She has run 21 miles already. Let t be the number of additional miles added on.

In total, she has run t+21 miles

This is going to be set greater than 38 since "Ann will run more than 38 miles"

So we have this inequality

t+21 > 38

we solve for t by subtracting 21 from both sides

t+21 > 38
t+21-21 > 38-21
t+0 > 17
t > 17

The final answer is t > 17

which means that the possible additional number of miles she could run is anything larger than 17. So t = 18 is one possibility.

8 0

3 years ago

Is 2 rational ? And why

zubka84 [21]

Yes, because it can be expressed as the quotient or fraction of two integers you

6 0

3 years ago

Read 2 more answers

More Algebra hw-- :/

grin007 [14]

Answer:
-11x=-66
x=6

Explanation:
-11x-12=-78
1) Add 12 to both sides.
-11x=-78+12
-11x=-66
2) Divide both sides by -11.
x=-66/-11
x=6

Hope this helps! Please leave a thanks and brainliest answer if you can :)

3 0

2 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$

$S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\$

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73$

5 0

3 years ago