Answer:
7 or 
Step-by-step explanation:
Each triangle has the surface area of 1.5, because 1 
3 = 3 / 2 = 1.5. You divide by two because it is half a rectangle instead of a full one. You multiply by four which would get you 1.5 4 = 6. The square in the middle has the surface area of 1 because 1 1 = 1. So you would get 7.
Rectangular area = l*b = 1620* 68 = 110160 cm² = 11.016 m²
the yearly increase of x% assumes is compounding yearly, so let's use that.
![95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r](https://tex.z-dn.net/?f=95000%3D80000%5Cleft%281%2B%5Cfrac%7B~~%20%5Cfrac%7Br%7D%7B100%7D~~%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%205%7D%5Cimplies%20%5Ccfrac%7B95000%7D%7B80000%7D%3D%5Cleft%28%201%2B%5Ccfrac%7Br%7D%7B100%7D%20%5Cright%29%5E5%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B19%7D%7B16%7D%3D%5Cleft%28%201%2B%5Ccfrac%7Br%7D%7B100%7D%20%5Cright%29%5E5%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D1%2B%5Ccfrac%7Br%7D%7B100%7D%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D%5Ccfrac%7B100%2Br%7D%7B100%7D%20%5C%5C%5C%5C%5C%5C%20100%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D100%2Br%5Cimplies%20100%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D-100%3Dr%5Cimplies%203.5%5Capprox%20r)
48 us fluid ounces =
3 us liquid pints
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.
