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KiRa [710]
3 years ago
12

A ladder reaches a window 20 feet above the ground. The foot of the ladder is 4 feet from the wall.

Mathematics
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

\sqrt{416} ft

or about 20.396ft when rounded to the third decimal place

Step-by-step explanation:

(assuming the question is to find the ladder's length)

let "l" be the ladder's length

20^{2} + 4^{2} = l^{2}

400 + 16 = l^{2} = 416

l = \sqrt{416} ft

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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
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Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



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