Given:
Area = 150 square feet
Bases = 14 feet and 16 feet
Find the height of the trapezoid.
Let us use the formula of finding the Area of a trapezoid to get the value of its height.
A = (a+b)/2 * h
150 ft² = (14ft + 16ft)/2 * h
150 ft² = 15ft * h
150 ft² / 15 ft = h
10 ft = h
The height of the trapezoid is 10 feet.
Let assume
X for ice cream cones
Y for sundaes
X+Y=172
X=172-Y……..*
1.10X+2.35Y=294.20
Substitute * in the above equation
1.10(172-Y)+2.35Y=294.20
189.2-1.10Y+2.35Y=294.20
1.25Y=294.20-189.2
1.25Y=105 divide both sides by 1.25
Y=84
The substitute Y value in * equation
X=172-84
X=88
Therefore
88 ice cream cones and 84 sundaes were sold!!
Precision refers to the degree of significant figures, or basically how many numbers after the decimal. It <span>is the </span>number<span> of digits in a </span><span>number.
</span> 5.9 liters and 13.6 liters have the same precision (one number after the decimal).
<span>90 liters and 94 liters have the same precision (no decimals).
Correct answer: A and B, C and D</span>
hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.
![\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Ba%7D%7B%5Csqrt%7B3%7D%7D~%2C~%5Cstackrel%7Bb%7D%7B1i%7D%29%5Cqquad%20%5Cbegin%7Bcases%7D%20r%3D%26%5Csqrt%7B%28%5Csqrt%7B3%7D%29%5E2%2B1%5E2%7D%5C%5C%20%26%5Csqrt%7B3%2B1%7D%5C%5C%20%262%5C%5C%20%5Ctheta%20%3D%26tan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Cright%29%5C%5C%5C%5C%20%26tan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%20%5Cright%29%5C%5C%20%26%5Cfrac%7B%5Cpi%20%7D%7B6%7D%20%5Cend%7Bcases%7D~%5Chfill%20%5Cimplies%20~%5Chfill%202%5Cleft%5B%20cos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D)
![\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~%5Cdotfill%5C%5C%5C%5C%20%5Cqquad%20%5Ctextit%7Bpower%20of%20two%20complex%20numbers%7D%20%5C%5C%5C%5C%5C%20%5B%5Cquad%20r%5Bcos%28%5Ctheta%29%2Bisin%28%5Ctheta%29%5D%5Cquad%20%5D%5En%5Cimplies%20r%5En%5Bcos%28n%5Ccdot%20%5Ctheta%29%2Bisin%28n%5Ccdot%20%5Ctheta%29%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%202%5Cleft%5B%20cos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D%20%5Cright%5D%5E3%5Cimplies%202%5E3%5Cleft%5B%20cos%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%208%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%20%5Cright%5D~%5Chfill)
