<u>9x</u> = <u>3y + 5</u>
9 9
x = 1/3y + 5/9
x - 5/9 = 1/3y + 5/9 - 5/9
x - 5/9 = 1/3y
3(x - 5/9) = 1/3y · 3
3x - 1 2/3 = y
y = 3x - 1 2/3
Answer:
Average speed is 37.35 mi/h
Step-by-step explanation:
given data
leave = 34 minutes before
church distance = 12.0 miles
average speed first 17 minutes = 5.0 mi/h
solution
so we find Total distance travel in first 17 minutes = speed × time
Total distance travel in first 17 minutes = 5 × 
Total distance travel in first 17 minutes = 1.416 mi
and
Distance Remaining = 12 - 1.416 = 10.584 mi
Time Remaining = 34 - 17 min = 17 min
so
remaining distance Average speed =
Average speed =
Average speed is 37.35 mi/h
Answer:
45+20=65
21+33=54
90+10=100
100+30= 130
Step-by-step explanation:
45+20=65
Doesn't belong as it has an odd numbered answer where as all the others have even answers
HOPE THIS HELPS
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS =
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²