see the attached figure to better understand the problem
we know that
in the right triangle ABC
cos 56°=AC/AB
where
AC is the adjacent side to angle 56 degrees------> the distance from the surveyor to the building
AB is the hypotenuse-----> 148 ft 2 in
56 degrees------> is the angle of elevation
so
cos 56°=AC/AB---------> solve for AC
AC=AB*cos 56°
AB=148 ft 2 in
convert 2 in to ft
1 ft -----> 12 in
x ft------> 2 in
x=2/12-----> x=0.17 ft
AB=148 ft 2 in-----> 148 ft+0.17 ft------> AB=148.17 ft
AC=AB*cos 56°----> AC=148.17*cos 56°------> AC=82.86 ft
convert 0.86 ft to in
0.86 ft=0.86*12-----> 10.32 in
distance AB=82 ft 10 in
the answer is
the distance from the surveyor to the building is 82 ft 10 in
Yes, people need to put the answers. Otherwise, it is just pointless to even ask your question.
Let the width path be x.
Length of the outer rectangle = 26 + 2x.
Width of the outer rectangle = 8 +2x.
Combined Area = (2x + 26)*(2x + 8) = 1008
2x*(2x + 8) + 26*(2x + 8 ) = 1008
4x² + 16x + 52x + 208 = 1008
4x² + 68x + 208 - 1008 = 0
4x² + 68x - 800 = 0. Divide through by 4.
x² + 17x - 200 = 0 . This is a quadratic equation.
Multiply first and last coefficients: 1*-200 = -200
We look for two numbers that multiply to give -200, and add to give +17
Those two numbers are 25 and -8.
Check: 25*-8 = -200 25 + -8 = 17
We replace the middle term of +17x in the quadratic expression with 25x -8x
x² +17x - 200 = 0
x² + 25x - 8x - 200 = 0
x(x + 25) - 8(x + 25) = 0
(x+25)(x -8) = 0
x + 25 = 0 or x - 8 = 0
x = 0 -25 x = 0 + 8
x = -25 x = 8
The width of the path can not be negative.
The only valid solution is x = 8.
The width of the path is 8 meters.
let's take admission fee= z
8z+11z=17.50+21.25
19z=38.75
z=38.25/19
z=2.4 (aprox)
Or In case of Kiara 17.50 amount paid in 8 rides
.so 1 ride=17.50/8
=2.1875
same In case of mia
so 1 ride=21.25/11
=1.9318.
but, they are not match so there is problem in your Questions.
Answer:
1.0.003
2.40 minutes
3.a,e
4.a,c,d
5.1/24
Step-by-step explanation:
