Answer:
QH = 227.8 km ≅ 228 km
Step-by-step explanation:
∵ The bearing from H to P is 084°
∵ The bearing from P to Q is 210°
∵ The distance from H to P = 340 km
∵ The distance from P to Q = 160 km
∴ The angle between 340 and 160 = 360 - 210 - (180 - 84) = 54°
( 180 - 84) ⇒ interior supplementary
By using cos Rule:
(QH)² = (PH)² + (PQ)² - 2(PH)(PQ)cos∠HPQ
(QH)² = 340² + 160² - 2(340)(160)cos(54) = 51904.965
∴ QH = 227.8 km ≅ 228 km
The inequality would be 34 > 2 x 3, if I'm correct. I'm not to sure on the graph part though , sorry.
Answer:
C. 5 2/3
Step-by-step explanation:
So what you do is see all the denomenators and factor them (denomenaotrs are the bottom numbres)
10,4,10,25
10=2 times 5
4=2 times 2
10=2 times 5
25=5 times 5
we need to include all of them so we need at leas two 2's, and two 5's
2 times 2 times 5 times 5=100
leact common denomator is 100
Step-by-step explanation:
the table is non-existent