3.7 and 4.7 because you are just adding 0.5 each time.
Well it depends on what the model is but if it's an IRA or whatever so if you make your own model that's easy
Answer:
7.00+2.50x>48.00
Step-by-step explanation:
He cant spend more than 48.00 so he'll have to spend less than or equal to 48.00
Check the picture below.
now, we have a triangle with all three sides, thus we can use Heron's Area Formula on the triangle.
![\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=10\\ b=26.695\\ c=22\\ s=29.3475 \end{cases} \\\\\\ A=\sqrt{29.3475(29.3475-10)(29.3475-26.695)(29.3475-22)} \\\\\\ A=\sqrt{29.3475(19.3475)(2.6525)(7.3475)}\implies A\approx \sqrt{11066.007} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill A\approx 105.195~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D10%5C%5C%20b%3D26.695%5C%5C%20c%3D22%5C%5C%20s%3D29.3475%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B29.3475%2829.3475-10%29%2829.3475-26.695%29%2829.3475-22%29%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B29.3475%2819.3475%29%282.6525%29%287.3475%29%7D%5Cimplies%20A%5Capprox%20%5Csqrt%7B11066.007%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20A%5Capprox%20105.195~%5Chfill)
13x-9=16
+9 +9
13x= 25
---- -----
13x 13x
x=1.92
Add nine to both sides of equation to cancel out the negative nine. Then, divide both sides by 13, which will result in the answer listed above.
