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Svetllana [295]
3 years ago
5

You place a 16 foot ladder against a building. The angle that the ladder forms with

Mathematics
1 answer:
gulaghasi [49]3 years ago
7 0

Answer:The base of the ladder should be placed so that it is one foot away from the building for every four feet of hight to where the ladder rests against the building. This is known as the 4 to 1 rule.

Step-by-step explanation:

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Step-by-step explanation:

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Answer:

X=1 and X=3

Step-by-step explanation:

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3x^2 - 4x = -2<br> Solve
amid [387]

Answer:

x =   \bigg \{\frac{ 2   - \sqrt{2} \: i }{3}, \:  \: \frac{ 2    +  \sqrt{2} \: i }{3} \bigg \}

Step-by-step explanation:

3 {x}^{2}  - 4x =  - 2 \\ 3 {x}^{2}  - 4x + 2 = 0 \\ equating \: it \: with  \\ a {x}^{2}  + bx + c = 0 \\ a = 3 \:  \: b =  - 4 \:  \: c = 2 \\  {b}^{2}  - 4ac \\  =  {( - 4)}^{2}  - 4 \times 3 \times 2 \\  = 16 - 24 \\  =  - 8  \\  \ {b}^{2}  - 4ac < 0  \\  \therefore \: given \: quadratic \: equation \: have \:  \\ imaginary \: solutios. \\  \\ x =  \frac{ - b \pm \sqrt{{b}^{2}  - 4ac } }{2a}  \\ =  \frac{ - ( - 4) \pm \sqrt{ - 8} }{2 \times 3} \\ =  \frac{ 4 \pm 2\sqrt{2} \: i }{2 \times 3}  \\  =  \frac{ 2 \pm \sqrt{2} \: i }{3} \\  \therefore \: x  = \frac{ 2   - \sqrt{2} \: i }{3}  \: or \: x  = \frac{ 2   +  \sqrt{2} \: i }{3} \:  \\  \\ x =   \bigg \{\frac{ 2   - \sqrt{2} \: i }{3}, \:  \: \frac{ 2    +  \sqrt{2} \: i }{3} \bigg \}

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