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IRISSAK [1]
3 years ago
10

Find by completing the square

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0
y=ax^2+bx+c\\\\y=a(x-h)^2+k\ where\ h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[tex]y=-3x^2+7x-5\\\\a=-3;\ b=7;\ c=-5\\\\h=\dfrac{-7}{2\cdot(-3)}=\dfrac{7}{6}\\\\k=\dfrac{-(7^2-4\cdot(-3)\cdot(-5)}{4\cdot(-3)}=\dfrac{-49+60}{-12}=-\dfrac{11}{12}\\\\y=-3\left(x-\dfrac{7}{6}\right)^2-\dfrac{11}{12}


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Help solve for x please (ignore my incorrect math)
IgorLugansk [536]

140° + ∠5 = 180°    <em>same side interior angles    </em>⇒  ∠5 = 40°

∠2 ≅ ∠5    <em>vertical angles   ⇒ </em>∠2 = 40°

∠8 = 53°   <em>alternate interior angles</em>

∠3 ≅ ∠8   <em>vertical angles  ⇒ </em>∠3 = 53°

x + ∠2 + ∠3 = 180°  <em>triangle sum theorem</em>

x + 40° + 53° = 180°

x + 93° = 180°

x = 87°

Answer: 87°

Note: I just realized that there is an easier way to solve this:

∠9 + 53° + x = 180°   <em>triangle sum theorem</em>

140° + ∠9 = 180°  <em>supplementary angles </em>⇒  ∠9 = 40°

40° + 53° + x = 180°  

x + 93° = 180°

x = 87°

Use whichever solutions makes sense to you.


     

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