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2 years ago

The slopes of perpendicular lines _____.

Mathematics

2 answers:

Annette [7]2 years ago

8 0

The answer is c.
their product is equal to -1

HACTEHA [7]2 years ago

7 0

Hey there

The slopes of perpendicular lines are negative reciprocals. Perpendicular lines have slopes that are negative reciprocals of each other.

= (C)

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(x + 1) (x+8) multiply binomials and put in standard form.

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Answer:

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HELP ASAP!!! explanation requiredd!

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(7x-5)+(3x-15)=180°

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
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Another way is apply the formula A=1/2*base*altitude,
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3. Let x be the circumference of the circle.

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4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

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1/3 + 5/6 + 5/12 What is the answer

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1/3 + 5/6 + 5/12

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