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3 years ago

5/9 - 3/8=? then estimate the answer.

Mathematics

1 answer:

Pani-rosa [81]3 years ago

8 0

$\frac{40}{72} - \frac{27}{72}$
$40 - 27 = 13$
$\frac{13}{72}$
$\frac{5}{9} - \frac{3}{8} = \frac{13}{72}$

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When given the two points (2,-3) and (2,-4) how do you write an equation in standard form

Nadusha1986 [10]

Answer:

2,-4

-2,-3

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Step-by-step explanation:

That is how write the equation because you have to line them up by the higher x or y value

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2 years ago

-12+x-34=60<br> If you like this question please rate this five stars!<br> Thank you!

natulia [17]

Answer:

x=106

Step-by-step explanation:

Let's solve your equation step-by-step.

−12+x−34=60

Step 1: Simplify both sides of the equation.

−12+x−34=60

−12+x+−34=60

(x)+(−12+−34)=60(Combine Like Terms)

x+−46=60

x−46=60

Step 2: Add 46 to both sides.

x−46+46=60+46

x=106

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3 years ago

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= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

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

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

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

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

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

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3 years ago

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Nonamiya [84]

Answer:

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3 years ago

Question Navigator

MArishka [77]

C and B are correct,