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AysviL [449]
3 years ago
6

The farthest distance a satellite signal can directly reach is the length of the segment tangent to the curve of Earth's surface

. If the measure of the angle formed by the tangent satellite signals is 137, what is the measure of the intercepted arc on Earth?
Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0
The measure of the intercepted arc on Earth is 43
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Step-by-step explanation:

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Step-by-step explanation:

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Help with question 5a
Llana [10]
This is a refreshing question!

We are given that
f(r)=ar+b, and
Sum f(r) =125 for r=1 to 5
Sum f(r) = 475 for r=1 to 10.

and we know, using Gauss's method, that
G(n)=sum (1,2,3.....n) = n(n+1)/2 or
G(n)=n(n+1)/2

Sum f(r) =125 for r=1 to 5
=>
sum=a(sum of 1 to 5) + 5b => G(5)a+5b=125   [G(5)=15]
15a+5b=125 ...................................................(1)

Similarly, Sum f(r) =  475 for r=1 to 10 => G(10)a+5b=475  [G(10)=55]
=>
55a+10b=475.................................................(2)

Solve system of equations (1) and (2)
(2)-2(1)
55-2(15)a=475-2(125) => 25a=225 =>
a=9

Substitute a=9 in 1 => 15(9)+5b=125 => 5b=-10
b=-2

Substitute a and b into f(r), 
f(r)=9r-2
check: sum f(r), r=1,5 = (9-2)+(18-2)+(27-2)+(36-2)+(45-2)=135-10=125 [good]

We define the sum of f(r) for r=1 to n as
S(n)=sum f(r) for r=1 to n = 9(sum 1,2,3....n)-2n = 9n(n+1)/2-2n = 9G(n)-2n
S(n)=9n(n+1)/2-2n
checks:
S(5)=9(15)-2(5)=135-10=125  [good]
S(10)=9(55)-2(10)=495-20=475 [good]

Hence 
(a)
S(n)=sum f(r) for r=1,n
= 9(sum i=1,n)+n(-2)
= 9(n(n+1)/2 -2n
=(9(n^2+n)/2) -2n

(b) sum f(r) for i=8,18
=sum f(r) for i=1,18  -  sum f(r) for i=1,7
=S(18)-S(7)
=(9(18^2-18)/2-2(18))-(9(7^2-7)/2-2(7))
=1503-238
=1265
5 0
2 years ago
PLEASE, I NEED HELP FOR A CLASS. The length of triangle base is 26. A line, which is parallel to the base divides the triangle i
sasho [114]

Answer:

DE=18.38

Step-by-step explanation:

It is given that the length of triangle base is 26 that is BC=26.

A line, which is parallel to the base divides the triangle into two equal area parts.

Therefore, from the given information, \frac{{\triangle}ADE}{ar ABC}=\frac{1}{2}.

Now, since it is given that A line, which is parallel to the base divides the triangle into two equal area parts, thus

\frac{ar ADE}{arABC}=\frac{1}{2}=\frac{(DE)^{2}}{(BC)^{2}}

⇒\frac{(DE)^{2}}{(BC)^{2}}=\frac{1}{2}

⇒\frac{(DE)^{2}}{(26)^{2}}=\frac{1}{2}

⇒(DE)^{2}=13{\times}26=338

⇒DE=18.38

7 0
3 years ago
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