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krek1111 [17]
3 years ago
12

In ΔPQR, \overline{PR} PR is extended through point R to point S, \text{m}\angle PQR = (2x+1)^{\circ}m∠PQR=(2x+1) ∘ , \text{m}\a

ngle QRS = (10x-10)^{\circ}m∠QRS=(10x−10) ∘ , and \text{m}\angle RPQ = (3x+14)^{\circ}m∠RPQ=(3x+14) ∘ . What is the value of x?X?
Mathematics
1 answer:
Vesna [10]3 years ago
6 0

Answer:

x = 5

Step-by-step explanation:

In ΔPQR, PR is extended through point R to point S.

m∠PQR=(2x+1) ∘

m∠QRS=(10x−10) ∘

m∠RPQ=(3x+14) ∘

Hence, we solve using Exterior angle Theorem.

This means that:

m∠QRS = m∠PQR + m∠RPQ

(10x - 10)° = (2x + 1)° + (3x + 14)°

10x - 10 = 2x + 1 + 3x + 14

Collect like terms

10x - 2x - 3x = 1 + 14 + 10

5x = 25

x = 25/5

x = 5

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x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

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x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21

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x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25ensiah193

4 0
3 years ago
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