hectictar

Let the number of unripe oranges  =  U

Let the number of ripe oranges  =  R

There were 18 more ripe oranges than unripe oranges. In other words,

the number of ripe oranges  =  the number of unripe oranges + 18

R  =  U + 18

The number of rotten oranges  =  24

84 – 24  =  60

R + U  =  60

Since  R  =  U + 18  we can substitute  U + 18  in for  R:

R + U  =  60

U + 18 + U  =  60

U + U  =  60 – 18

2U  =  42

U  =  21

Now that we know the value of  U  we can find the value of  R

R  =  U + 18

R  =  21 + 18

R  =  39

the number of ripe oranges  =  39

the number of rotten oranges  =  24

the number of unripe oranges  =  21

num of ripe oranges : num of rotten oranges : num of unripe oranges   =   39 : 24 : 21   =   13 : 8 : 7

Let Mai's height  =  m

Let Jon's height  =  j

We want to find  \(\frac{m}{j}\)

\(\frac14\cdot m\ =\ \frac25\cdot j\)

                             Multiply both sides of the equation by  4

\(m\ =\ \frac85\cdot j\)

                             Divide both sides of the equation by  j

\(\frac{m}{j}\ =\ \frac85\)

The ratio of  m  to  j  is  \(\frac85\)

Draw a height from the base to create two congruent 30-60-90 triangles.

the side across from the  60°  angle  =  10

the side across from the  30°  angle  =  10/√3

area of isosceles triangle  =  ( 1/2 )( base )( height )

area of isosceles triangle  =  ( 1/2 )( 20 ft )( 10/√3 ft )

area of isosceles triangle  =  100 / √3     square feet

area of isosceles triangle  =  100√3 / 3   square feet

Here's equilateral triangle ABC and a circle with radius 1 centered on point A:

The intersecion of the triangle and the circle is the highlighted sector.

probability that a randomly chosen point lands in highlighted sector  =  area of sector / area of triangle

So we just have to find the area of the sector and the area of the triangle.

Let's find the area of the sector:

\(\frac{\text{area of sector}}{\text{area of circle}}\ =\ \frac{\text{measure of central angle}}{360^\circ} \\~\\ \frac{\text{area of sector}}{\pi\cdot1^2}\ =\ \frac{60^\circ}{360^\circ} \\~\\ \text{area of sector}\ =\ \frac{60^\circ}{360^\circ}\cdot\pi\cdot1^2 \\~\\ \text{area of sector}\ =\ \frac{\pi}{6} \)

Now let's find the area of the triangle:

\(\text{area of triangle}\ =\ \frac12\cdot\text{base}\cdot\text{height}\\~\\ \text{area of triangle}\ =\ \frac12\cdot3\cdot\frac{3\sqrt3}{2}\\~\\ \text{area of triangle}\ =\ \frac{9\sqrt3}{4}\)

Now we can find the probability in question.

\(\text{probability} = \frac{\text{area of sector} }{ \text{area of triangle}}\\~\\ \text{probability} = \text{area of sector} \div \text{area of triangle} \\~\\ \text{probability} = \frac{\pi}{6} \div \frac{9\sqrt3}{4} \\~\\ \text{probability} = \frac{\pi}{6} \cdot \frac{4}{9\sqrt3} \\~\\ \text{probability} = \frac{4\pi}{54\sqrt3} \\~\\ \text{probability} = \frac{4\sqrt3\pi}{162} \\~\\ \text{probability} = \frac{2\sqrt3\pi}{81} \)_

\((4-a)^3\ =\ 32\)

                                         Take the cube root of both sides of the equation.

\(\sqrt[3]{(4-a)^3}\ =\ \sqrt[3]{32}\)

                                         Simplify the left side with the rule  \(\sqrt[3]{n^3}\ =\ n\)

\(4-a\ =\ \sqrt[3]{32}\)

                                                    We can rewrite  32  like this because  32 = 2 * 2 * 2 * 2 * 2

\(4-a\ =\ \sqrt[3]{2\cdot2\cdot2\cdot2\cdot2}\)

                                                    We can rewrite the right side again like this...

\(4-a\ =\ \sqrt[3]{2\cdot2\cdot2}\cdot\sqrt[3]{2\cdot2}\)

                                                    And   2 * 2 * 2  =  2³   and   2 * 2  =  4

\(4-a\ =\ \sqrt[3]{2^3}\,\cdot\,\sqrt[3]{4}\)

                                                    Simplify  \(\sqrt[3]{2^3}\)  again with the rule  \(\sqrt[3]{n^3}\ =\ n\)

\(4-a\ =\ 2\,\cdot\,\sqrt[3]{4}\)

\(4-a\ =\ 2\sqrt[3]{4}\)

                               Add  a  to both sides of the equation.

\(4\ =\ 2\sqrt[3]{4}+a\)

                               Subtract  \(2\sqrt[3]{4}\)  from both sides of the equation.

\(4-2\sqrt[3]{4}\ =\ a\)

\(a\ =\ 4-2\sqrt[3]{4}\)-

Until this gets fixed, you should be able to use another service to upload your image.

We can find the length of the missing sides using the information that the area is 240 sq units. 

area of triangle  =  ( 1/2 )( base )( height )     Plug in the information we're given.

240  =  ( 1/2 )( 20 )( height )     Multiply  1/2   by   20

240  =  ( 10 )( height )          Divide both sides of the equation by  10

24  =  height

So we know the base is  20  and the height is  24

When we draw a height to the base, we split the triangle into two smaller congruent triangles.

(We can know they're congruent by the hypotenuse-leg congruence theorem.)

And so the height splits the base exactly in half.

That means the length of the missing leg in each of the right triangles  =  20/2  =  10

So by the Pythagorean Theorem,

10² + 24²  =  c²

100 + 576  =  c²

676  =  c²

c  =  √676

c  =  26

Now do you see how to finish it?

{ a^-1 }  =  { a² }

Using the definition of the  { }  function, we can say

a^-1 - floor( a^-1 )  =  a² - floor( a² )

Now since  2 < a² < 3 ,  a  can't be  1 .  And so  floor( a^-1 )  =  0

Also since  2 < a² < 3 ,  we can say for sure that  floor( a² )  =  2

a^-1 - 0  =  a² - 2

                              Subtract  a^-1  from both sides of the equation.

0  =  a² - 2 - a^-1

                              Multiply through by  a  and note  a ≠ 0

0  =  a³ - 2a - 1

0  =  a³ - a  -  a - 1

0  =  a³ - a² - a  +  a² - a - 1

0  =  a(a² - a - 1) + 1(a² - a - 1)

0  =  (a + 1)(a² - a - 1)

We only want the middle solution because the other two violate the inequality  2 < a² < 3

So     a  =  (1 + √5 ) / 2

An inscribed angle is half the measure of its intercepted arc, so

m∠ADC  =  (1/2)(m∠ABC)

                                              Substitute  23°  in for  m∠ADC

23°  =  (1/2)(m∠ABC)

                                              Multiply both sides by  2

46°  =  m∠ABC

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