3 years ago

Help please i need it dhhdkhdkjhdkdhkshdksj

Mathematics

1 answer:

kakasveta [241]3 years ago

8 0

Answer:

26 percent sodium

also can i have brainliest please

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Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched fr

Andreyy89

Answer:

Step-by-step explanation:

Given

height of two Poles are 60 and 80 ft

Distance between them is 100 ft

Let x be the distance of Pole of ht 80 ft from Point of stretch

thus length of rope is given by

$L=L_1+L_2$

$L_1=\sqrt{80^2+x^2}$

$L_2=\sqrt{60^2+(100-x)^2}$

$L=\sqrt{80^2+x^2}+\sqrt{60^2+(100-x)^2}$

differentiate w.r.t x we get

$\frac{\mathrm{d} L}{\mathrm{d} x}=\frac{2x}{2\sqrt{80^2+x^2}}-\frac{2\left ( 100-x\right )}{\sqrt{60^2+\left ( 100-2x\right )^2}}$

Put $\frac{\mathrm{d} L}{\mathrm{d} x}=0$ to get minimum value

$\frac{2x}{2\sqrt{80^2+x^2}}=\frac{2\left ( 100-x\right )}{\sqrt{60^2+\left ( 100-2x\right )^2}}$

squaring

$x^2\left ( (100-x)^2+60^2\right )=(100-x)^2(80^2+x^2)$

Rearranging

$28x^2-1800x+640000=0$

$x=\frac{400}{7}$

thus $L_1=98.312 ft$

$L_2=73.73 ft$

$L=172.04 ft$

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3 years ago