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nasty-shy [4]
3 years ago
10

A 18 ft tall flagpole standing next to a tent costs o 27 ft shadow. If the tent casts a shadow that is 15 ft long. then how tall

is it?
Mathematics
2 answers:
Eva8 [605]3 years ago
8 0
24 feet tall
hope it’s correct :)
spin [16.1K]3 years ago
6 0
The tent would be 24 feet tall
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3x to the power of 3 + 21x squared + 36x = 0 by factoring the quadratic equation.
Diano4ka-milaya [45]

Answer:

x = 0, -4 and -3.

Step-by-step explanation:

The given expression is "3x to the power of 3 + 21x squared + 36x = 0".

We can write this expression as follows :

3x^3+21x^2+36x=0

Taking 3x common.

3x(x^2+7x+12)=0\\\\3x=0\ \text{and}\ (x^2+7x+12)=0

x = 3

x^2+7x+12=0\\\\x^2+3x+4x+12=0\\\\x(x+3)+4(x+3)=0\\\\(x+4)(x+3)=0\\\\x=-4,-3

Hence, the solution of the given quadratic equation are x = 0, -4 and -3.

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RideAnS [48]

\sf{Given : 3tanx + 7 = \dfrac{2}{(1 - sinx)(1 + sinx)}}

We know that : (a - b)(a + b) = a² - b²

\implies \sf{3tanx + 7 = \dfrac{2}{1 - sin^2x}}

We know that : 1 - sin²x = cos²x

\implies \sf{3tanx + 7 = \dfrac{2}{cos^2x}}

\sf{\bigstar \ \ We \ know \ that : \boxed{\sf{\dfrac{1}{cos^2x} = sec^2x}}}

\implies \sf{3tanx + 7 = 2sec^2x}

We know that : sec²x = 1 + tan²x

\implies \sf{3tanx + 7 =2(1 + tan^2x)}

\implies \sf{2 + 2tan^2x - 3 tanx - 7 = 0}

\implies \sf{2tan^2x - 3 tanx - 5 = 0}

\implies \sf{2tan^2x -  5tanx + 2tanx - 5 = 0}

\implies \sf{2tanx(tanx + 1) - 5(tanx + 1) = 0}

\implies \sf{(tanx + 1)(2tanx - 5) = 0}

\implies \sf{tanx = -1 \ (or) \ tanx = \dfrac{5}{2} }

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3 years ago
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