Let
x--------> <span>the speed of the boat in mph
we know that
</span>going upstream the speed is x-7
<span>going downstream the speed is x+7
</span>
let the distance be d
<span>time to go upstream = d/(x-7) </span>
<span>time to go downstream = d/(x+7)
</span>
time going upstream is 3 times going downstream
so
<span>3d/(x+7) = d/(x-7) </span>
<span>divide by d </span>
<span>3/(x+7) = 1/(x-7) </span>
<span>3x-21 = x+7 </span>
<span>2x = 28 </span>
<span>x = 14
</span>
the answer is
<span>the speed of the boat in still water is 14 mph</span>
The domain and simplified terms are marked below
Answer:
The answer to this question is x=2
We know that Sum of Angles in a Triangle is Equal to 180°
Here EBF is a Triangle
⇒ m∠EBF + m∠BEF + m∠EFB = 180°
⇒ 60° + 40° + m∠EFB = 180°
⇒ 100° + m∠EFB = 180°
⇒ m∠EFB = 180° - 100°
⇒ m∠EFB = 80°
As Line m and Line p are Parallel Lines :
Alternate Interior Angles are Equal, here Alternate Interior Angles are m∠BEF and m∠ABE
⇒ m∠BEF = m∠ABE
⇒ m∠ABE = 40°
We know that Vertically Opposite Angles are Equal, Here m∠GFI and m∠EFB are Vertically Opposite Angles.
⇒ m∠GFI = m∠EFB
⇒ m∠GFI = 80°
We can notice that m∠DEB and m∠BEF form a Linear Pair
⇒ m∠DEB + m∠BEF = 180°
⇒ m∠DEB + 40° = 180°
⇒ m∠DEB = 180° - 40°
⇒ m∠DEB = 140°
We can notice that Sum of Angles m∠CBF and m∠EBF and m∠ABE is 180°
⇒ m∠CBF + m∠EBF + m∠ABE = 180°
⇒ m∠CBF + 60° + 40° = 180°
⇒ m∠CBF + 100° = 180°
⇒ m∠CBF = 180° - 100°
⇒ m∠CBF = 80°
We can notice that m∠BFG and m∠EFB form a Linear Pair
⇒ m∠BFG + m∠EFB = 180°
⇒ m∠BFG + 80° = 180°
⇒ m∠BFG = 180° - 80°
⇒ m∠BFG = 100°
We know that Vertically Opposite Angles are Equal, Here m∠BFG and m∠IFE are Vertically Opposite Angles.
⇒ m∠BFG = m∠IFE
⇒ m∠IFE = 100°
