Where’s the cone? Is there a picture of a cone? I can’t really solve this, I’m sorry if I bother you
Answer:
1. D
2. B
3. A
Step-by-step explanation:
Question 1:
The pair of <JKL and <LKM can be referred to as linear pairs. They are two adjacent angles that are formed from the intersecting of two lines.
Question 2:
Given that <KLM = x°
<KML = 50°
<JKL = (2x - 15)°
According to the exterior angle theorem, exterior ∠ JKL = <KLM + KML.
2x - 15 = x + 50
Solve for x
2x - x = 15 + 50
x = 65
Therefore, <KLM = 65°
QUESTION 3:
<JKL = 2x - 15
Plug in the value of x
<JKL = 2(65) - 15
= 130 - 15
<JKL = 115°
Answer:
312 copies
Step-by-step explanation:
100 + 100 = 200
40 + 60 = 100
8 + 4 = 12
200 + 100 + 12 = 312 copies
164+219=383 is exact for estimate round 164 to 165 or 160 and 219 to 220 so
165+220=385 or 160+220=380
Class A: 6v + 8b = 202
Class B: 12v + 16b = 284
Solve using the elimination method:
since 6v and 12v are perfect for elimination, multiply the class A equation by 2 so that the van variable cancels out:
12v + 16b = 404
12v + 10b = 284
Then subtract the bottom equation from the top:
6b = 120
b = 20
Now you know that each bus can hold 20 students.
Just plug this into one of the original equations to solve for vans:
6v + 8(20) = 202
6v + 160 = 202
6v = 42
v=7
So then you know that each van can hold 7 students.
Check:
12 (7) + 10 (20) = 284
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