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irga5000 [103]
3 years ago
13

Sky high. on a sunny, warm day, a student decides to fly a kite on the college green just to relax. his kite takes off and soars

. he lets all 150 feet of the string out and attracts a crowd of onlookers. there is a slight breeze, and a spectator 90 feet away from the student notices that the kite is directly above her. unlike a real kite, this math-question kite has the string going in a straight line from the student to the kite. how high is the kite from the ground?
Mathematics
1 answer:
Ahat [919]3 years ago
4 0
60 feet high is your answer
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What is the fourth term in the binomial expansion (a+b)^6)
Dafna11 [192]

Answer:

20a^3b^3

Step-by-step explanation:

<u>Binomial Series</u>

(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example:  4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3

\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3

\implies \dfrac{6!}{3!3!}a^{3}b^3

\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3

\implies \left(\dfrac{120}{6}\right)a^{3}b^3

\implies 20a^3b^3

7 0
2 years ago
A uniform distribution is defined over the interval from 6 to 10.
scoundrel [369]
Are you given an equation?
6 0
3 years ago
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Travka [436]

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Step-by-step explanation:

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4 0
3 years ago
Yesenia swims laps at a pool. For every 10 laps she swims the backstroke, she swims 15 laps of freestyle. Which ratio shows the
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Answer:

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8 0
3 years ago
How many solutions are in -x-y=3 and y=x-3
Klio2033 [76]
See, pls, offered explanation:
1) the both equations are linear function, it means system made up of the both equation has the only solution.
2) if to prove this:
\left \{ {{-x-y=3} \atop {-x+y=-3}} \right.  \left \{ {{x=0} \atop {y=-3}} \right.
Answer: one solution.
5 0
3 years ago
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