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bagirrra123 [75]

3 years ago

10

adam use the three fractions to make a circle graph and colored each a different color.What fraction of the graph is not colored

? the three fractions are 3/12 1/6 1/3

1 answer:

never [62]3 years ago

6 0

3/12+2/12+4/12= 9/12 12/12-9/12= 3/12 So 1/4 of the graph is not colored

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Simplify the expression please help

Mila [183]

Simplify the following:

((-1^3)/(-3)^(-3))^2

1^3 = 1:
((-1)/(-3)^(-3))^2

(-3)^(-3) = 1/(-1)^3×1/3^3 = (-1)/3^3:
((-1)/((-1)/3^3))^2

3^3 = 3×3^2:
((-1)/(-1/(3×3^2)))^2

3^2 = 9:
((-1)/((-1)/(3×9)))^2

3×9 = 27:
((-1)/((-1)/27))^2

Multiply the numerator of (-1)/((-1)/27) by the reciprocal of the denominator. (-1)/((-1)/27) = (-27)/(-1):
((-27)/(-1))^2

(-27)/(-1) = (-1)/(-1)×27 = 27:
27^2

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3 0

3 years ago

Hi can you guys please help me on these two?!

lesya692 [45]

Answer:

1. R and Z 2. AC and AE

Step-by-step explanation:

5 0

2 years ago

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REALLY NEED HELP: The Venn Diagram shows a universe with 26 elements and two different outcomes: B and Y

Goshia [24]

Answer:

P(B|Y)=0.4

Step-by-step explanation:

We are given with a Venn diagram of a universe with 26 elements and two different outcomes: B and Y

And we are asked to find P(B|Y)

The formula for P(B|Y)= $\frac{P(B\cap Y)}{P(Y)}$

From Venn diagram, $P(B\cap Y)=\frac{n(B\cap Y)}{26} =\frac{8}{26} =\frac{4}{13}$

And P(Y)= $\frac{n(Y)}{26}=\frac{8+12}{26}= \frac{20}{26}= \frac{10}{13}$

Hence P(B|Y)= $\frac{(\frac{4}{13})}{\frac{10}{13}} =\frac{4}{10}=0.4$

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3 years ago

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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= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
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,
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$2 \pi r=x$

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

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

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

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

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3 years ago

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grandymaker [24]

Answer:

it b

Step-by-step explanation:

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2 years ago