geometry help on inequalities and triangles?
in triangle stw st=15 and tw=19 write an inequality to show all possible values for sw
tw - st < sw < st + tw

4 < sw < 34
18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units hello, red.since the triangle is a right triangle, you can use the pythagorean theorem.let x be the length of the unknown side.then,92 + x2 = (9√2)281 + x2 = 81·2 = 162x2 = 162 - 81x2 = 81take the square root of both sides.x = ±9since the length cannot be a negative number, x = 9 m We are given:  

\(\displaystyle{t}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}\)  

Subtract t0 from both sides:  

\(\displaystyle{t}-{t}{0}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}-{t}{0}\)  

\(\displaystyle{t}-{t}{0}={\left({\frac{{{k}}}{{{2}}}}\right)}{w}\)  

Multiply both sides by \(\frac{2}{k}\)  

\(\displaystyle{\left({t}-{t}{0}\right)}{\frac{{{2}}}{{{k}}}}={\left({\frac{{{k}}}{{{2}}}}\right)}{w}\cdot{\left({\frac{{{2}}}{{{k}}}}\right)}\)  

\(\displaystyle{\frac{{{2}{\left({t}-{t}{0}\right)}}}{{{k}}}}={w}\)  

\(\displaystyle{w}={2}\frac{{{t}-{t}{0}}}{{k}}\) find and correct the errors in the following mathematical statements

q19) one idea can be videos have frames so, vectorize each frame and calculate dtw between two consecutive frames. this will help in gaining the difference in each frame and perhaps help in classification. hope this helps you. the distance to be covered to go one round along perimeter of this quadrilateral is12 cm + 3 cm + 5.6 cm + 19 cm let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved perhaps what you are looking for is in objective and nonlinear constraints in the same function.
alan weiss
matlab mathematical toolbox documentation ?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92