1/2in = 16 mi
2 3/4 in without whole numbers [(denominator ×whole) + numerator] / denominator
4×2 = 8 +3 = 11/4
Find common denominator for 1/2 and 11/4 (common D Is 4)
Multiply 1/2 by 2/2
2/4 in = 16 mi divide this by 2
2/4 ×1/2 = 1/4
16 /2 = 8
1/4 = 8
1/4 goes into 11/4 11 times
Multiply 8 times 11
88 miles (D)
End zone to end zone is = length so his scale is 6 in and in real life it is 360 ft
To find scale divide 360 by 6
1in = 60 ft (B)
160 ft wide
(divide 160 by 60 ft ) fraction
160/60 is a fraction now simplify
16/6
Common factors of both numbers is 2
Simplify by 2
8/3
Make into whole number fraction
3 goes into 8 two times (2×3=6)(2 becomes whole number) while there's is 2 left over( 8 -6=2) (becoming the numerator)
2 2/3 in (B)
120 ft long
80 ft wide
24 in is scale of "long"
Divide 120 by 24
120÷24=5
So 5in in scale equals 1ft in real life
To find width in scale divide 80 by 5
80/5= 16in (A)
If you have any trouble understanding feel free to let me know I will try to explain as best as I can :)
The rest of the question is the attached figure.
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17
The sum must be irrational in contradiction to it being rational
I forgot all about this :(