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kupik [55]
3 years ago
13

Help please!!! Find x and Y

Mathematics
1 answer:
Sav [38]3 years ago
8 0

Answer:

x = 40

y = 15

Step-by-step explanation:

Given is an equilateral triangle.

Measure of each angle of an equilateral triangle is 60°.

Therefore,

(x + 20)° = 60°

x + 20 = 60

x = 60 - 20

x = 40

4y = 60

y = 60/4

y = 15

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\large\boxed{Q1:\ x=2\ or\ x=5}\\\boxed{Q2:\ x=1-\sqrt{21}\ or\ x=1+\sqrt{21}}

Step-by-step explanation:

\text{Use the quadratic formula:}\\\\ax^2+bx+c=0\\\\\text{If}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\==========================================

\bold{Q1}\\\\x^2-7x+10=0\\\\a=1,\ b=-7,\ c=10\\\\b^2-4ac=(-7)^2-4(1)(10)=49-40=9>0\\\\\sqrt{b^2-4ac}=\sqrt9=3\\\\x_1=\dfrac{-(-7)-3}{2(1)}=\dfrac{7-3}{2}=\dfrac{4}{2}=2\\\\x_2=\dfrac{-(-7)+3}{2(1)}=\dfrac{7+3}{2}=\dfrac{10}{2}=5\\\\========================================

\bold{Q2}\\x^2-2x=20\qquad\text{subtract 20 from both sides}\\\\x^2-2x-20=0\\\\a=1,\ b=-2,\ c=-20\\\\b^2-4ac=(-2)^2-4(1)(-20)=4+80=84>0\\\\\sqrt{b^2-4ac}=\sqrt{84}=\sqrt{4\cdot21}=\sqrt4\cdot\sqrt{21}=2\sqrt{21}\\\\x_1=\dfrac{-(-2)-2\sqrt{21}}{2(1)}=\dfrac{2-2\sqrt{21}}{2}=1-\sqrt{21}\\\\x_2=\dfrac{-(-2)+2\sqrt{21}}{2(1)}=\dfrac{2+2\sqrt{21}}{2}=1+\sqrt{21}

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3 years ago
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