The quadratic formula is
and in the equation ax^2+bx+c=0
so now all you have to do is substitute the numbers into the quadratic formula
Answer:
The variable, y is 11°
Step-by-step explanation:
The given parameters are;
in triangle ΔABC; in triangle ΔFGH;
Segment = 14 Segment = 14
Segment = 27 Segment = 19
Segment = 19 Segment = 2·y + 5
∡A = 32° ∡G = 32°
∡A = ∠BAC which is the angle formed by segments = 14 and = 19
Therefore, segment = 27, is the segment opposite to ∡A = 32°
Similarly, ∡G = ∠FGH which is the angle formed by segments = 14 and = 19
Therefore, segment = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;
≅ by Congruent Parts of Congruent Triangles are Congruent (CPCTC)
∴ = = 27° y definition of congruency
= 2·y + 5 = 27° by transitive property
∴ 2·y + 5 = 27°
2·y = 27° - 5° = 22°
y = 22°/2 = 11°
The variable, y = 11°
Answer:
<em>Hey mate, here's ur answer:</em>
<em>---------------------------------------------------</em>
<em>-2x^5. 4x^2 </em>
<em>=(−2)*x5*4*x2
</em>
<em>=−8*x7
</em>
<u><em>=−8x7</em></u>
<em>--------------------------------------------------</em>
<em>Hope it helps</em>
<em>#stayhomestaysafemate</em>
<em>:D</em>
Answer:
Step-by-step explanation:
┈➤ . . ⇢ ˗ˏˋ⌦ ..♡ˎˊ˗ ꒰✈꒱
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★ . ꜝꜞ ᳝ ࣪ hope that helped! ʬʬ