Unfortunately there isn't enough information.
Check out the diagram below. We have segment BC equal to 120 meters long. Points B, C, D and E are all on the edge of the same circle. According to the inscribed angle theorem, angles BDC and BEC are congruent. This shows that the surveyor could be at points D or E, or the surveyor could be anywhere on the circle. There are infinitely many locations for the surveyor to be at, which leads to infinitely many possible widths of this canal.
Think of the 13-ft length of the ladder as the hypotenuse of a right triangle. Represent the horiz. distance from foot of ladder to base of tree by x, or 5 ft.
Represent the vert. dist. from base of tree to top of ladder by y, which is unknown.
Then (13 ft)^2 = (5 ft)^2 + y^2, or
169 ft^2 = 25 ft^2 + y^2. This simplifies to y^2 = 144. Thus y = + 12 feeet.
Note: Please pay attention to your spelling: "lader i up agenst a tree" should be "the top of a 13-ft ladder is placed against a tree."
Answer:
D. x=1
Step-by-step explanation:
Image can be rewritten as 3x + 6 = 9.
Solve for x.
3x +6 = 9
Isolate x by subtracting 6 from both sides.
3x = 3
Get rid of the 3 by dividing by 3 on both sides.
x = 1
Hope this helped.
Height covered by the plane=h= 500 feet
Distance covered by the plane along the ground=d= 2 mile
Now, convert the mile into feet,
1 mile = 5280 feet
2 mile = 10560 feet
Now, use the formula
h= d tanθ
tanθ= h/d
tanθ= 500/10560
tanθ= 25/528
Take the tan⁻¹ on both side,
θ= tan⁻¹(25/528)
θ =2.71°
0.38 if you divide 100 by 8 you get 0.125 and when you multiply it by 3 you get 0.375. 0.38>0.375
