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4 years ago

A triangle is rotated 90° about the orgin. Which rule describes the transformation

Mathematics

1 answer:

Katen [24]4 years ago

8 0

Step-by-step explanation:

If a triangle is rotated 90° about the origin.

The rule that describes the transformation is

(x, y) → (–y, x)

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Brenda is planning her birthday party. She wants to have 12.6 liters of punch, 10.5 gallons of ice cream, 3.9 pounds of fudge, a

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3 years ago

What is the length of a in this triangle?

gizmo_the_mogwai [7]

Use Pythagorean’s Theorem. a^2 + b^2=c^2.
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3 years ago

Which ones are the Hypotenuse?

kodGreya [7K]

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Step-by-step explanation:

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3 years ago

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

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