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djyliett [7]
3 years ago
10

Answer ASAP. Please give correct answer. (Involves a bunch of circles and trig identities)

Mathematics
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

45.40

Step-by-step explanation:

First of all, the shape of rope is not a parabola but a catenary, and all catenaries are similar, defined by:

y=acoshxa

You just have to figure out where the origin is (see picture). The hight of the lowest point on the rope is 20 and the pole is 50 meters high. So the end point must be a+(50−20) above the x-axis. In other words (d/2,a+30) must be a point on the catenary:

a+30=acoshd2a(1)

The lenght of the catenary is given by the following formula (which can be proved easily):

s=asinhx2a−asinhx1a

where x1,x2 are x-cooridanates of ending points. In our case:

80=2asinhd2a

40=asinhd2a(2)

You have to solve the system of two equations, (1) and (2), with two unknowns (a,d). It's fairly straightforward.

Square (1) and (2) and subtract. You will get:

(a+30)2−402=a2

Calculate a from this equation, replace that value into (1) or (2) to evaluate d.

My calculation:

a=353≈11.67

d=703arccosh257≈45.40

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Williams salary is £24000 his salary increases by 4% work out Williams new salary
Ad libitum [116K]
100%  = 24000 
50 %  =  12000
10%   =   2400
5 %    = 1200
1 %    = 240
5 0
3 years ago
The legs of a right triangle are 10 cm and 24 cm. What is the length of the hypotenuse?
n200080 [17]
To determine the length of the hypotenuse, apply Pythagorean theorem.

A^2 + B^2 = C^2
(10)^2 + (24)^2 = C^2
100 + 576 = C^2
676 = C^2
C = 26 cm.

The hypotenuse is 26 cm.
6 0
3 years ago
Read 2 more answers
Please help me thank you!
Romashka [77]

Answer:

the answer is 1¢26 if not try h5

7 0
2 years ago
So are you good at maths then what is
quester [9]

maybe 28 is the answer...

4 0
3 years ago
Read 2 more answers
PLZ HELP DONT UNDERSTAND!!!
victus00 [196]
\bf \textit{area of a triangle}=A=\cfrac{1}{2}bh\qquad 
\begin{cases}
b=base\\
h=\textit{height or altitude}
\end{cases}\\\\
-----------------------------\\\\
\textit{area of a trapezoid}\\\\
A=\cfrac{h}{2}(a+b)\qquad 
\begin{cases}
a=\textit{base, or one of the parallel sides}\\
b=\textit{base, or the other parallel side}\\
h=\textit{height, the distance between both}\\
\qquad \textit{parallel sides}
\end{cases}
7 0
3 years ago
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