Answer:
x = 65 , y = 64
Step-by-step explanation:
y - 16 and 2y + 4 are same- side interior angles and sum to 180° , that is
y - 16 + 2y + 4 = 180
3y - 12 = 180 ( add 12 to both sides )
3y = 192 ( divide both sides by 3 )
y = 64
then
2y + 4 = 2(64) + 4 = 128 + 4 = 132
the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.
2y + 4 is an exterior angle of the triangle , then
x + 67 = 132 ( subtract 67 from both sides )
x = 65
Answer:
120 degrees
Step-by-step explanation:
This is a quadrilateral, and by definition, the sum of all the interior angles is 360. So, we can just sum up all the expressions and set them equal to 360:
2x + x + 2x + x = 360
Combine like terms and solve for x:
6x = 360
x = 360/6 = 60
<BCD = 2x = 2 * 60 = 120
Thus, the answer is 120 degrees.
Hope this helps!
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]
Answer:
Yes they are
Step-by-step explanation:
In the triangle JKL, the sides can be calculated as following:
=> JK = 
=> JL = 
=> KL = 
In the triangle QNP, the sides can be calculate as following:
=> QN = ![\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-3-%28-4%29%5D%5E%7B2%7D%20%2B%20%280-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B1%5E%7B2%7D%2B%28-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B1%2B16%7D%3D%5Csqrt%7B17%7D)
=> QP = ![\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-7-%28-4%29%5D%5E%7B2%7D%20%2B%20%281-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-3%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B9%2B9%7D%3D%5Csqrt%7B18%7D%20%3D%203%5Csqrt%7B2%7D)
=> NP = ![\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-7-%28-3%29%5D%5E%7B2%7D%20%2B%20%281-0%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B%28-4%29%5E%7B2%7D%2B1%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B16%2B1%7D%3D%5Csqrt%7B17%7D)
It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP
=> They are congruent triangles
Answer:
Your teacher is right, there is not enough info
Step-by-step explanation:
<h3>Question 1</h3>
We can see that RS is divided by half
The PQ is not indicated as perpendicular to RS or RQ is not indicates same as QS
So P is not on the perpendicular bisector of RS
<h3>Question 2</h3>
We can see that PD⊥DE and PF⊥FE
There is no indication that PD = PF or ∠DEP ≅ FEP
So PE is not the angle bisector of ∠DEF
Answer:
(4,0)
Step-by-step explanation:
we have
----> inequality A
----> inequality B
we know that
If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)
Verify each ordered pair
case 1) (4,0)
<em>Inequality A</em>
----> is true
<em>Inequality B</em>

----> is true
so
the ordered pair makes both inequalities true
case 2) (1,2)
<em>Inequality A</em>
----> is not true
so
the ordered pair not makes both inequalities true
case 3) (0,4)
<em>Inequality A</em>
----> is not true
so
the ordered pair not makes both inequalities true
case 4) (2,1)
<em>Inequality A</em>
----> is true
<em>Inequality B</em>

----> is not true
so
the ordered pair not makes both inequalities true