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9 days / 432 pages    *   720 pages   =  15 days 

                      ( see how the 'pages' cancel' out .....and you are left with 'days' as units?)

Let x number of  how many time it take to read.

$\frac{9}{432}=\frac{x}{720}$

$48x=720$

$x=15$

Range : f(x)>= 0

432/9 = 48

720/48 = 15

15 days

The answer is 4 ONLY if quadrilateral EFGH is the parallelogram. (???)

Like so

$(7x+2)+(7x+2)+(180-(17x-23))=180$

$207-3x=180$

$3x=27$

$x=9$

Because $\bigtriangleup ABC\sim \bigtriangleup ADE$, then :

$\frac{AB}{BC}=\frac{AD}{DE}$

$\frac{2}{x}=\frac{3}{1}$

$3x=2$

$x=\frac{2}{3}$

Nice Answer James!
But, I would like to think of it a different way.

0.9 cannot be equal to 1 only $0.\overline{9}$ can. If you are meaning$0.\overline{9}$ then, I can show you how to do it.

1/9 = $0.\overline{1}$

2/9 = $0.\overline{2}$

3/9 = 1/3 = $0.\overline{3}$

And so on this means that when we come to 9/9 = $0.\overline{9}$=1

🐺🐺🐺

We know that x-3 = 4x-15 so we can just factorize I think. 

1.x-3 = 4x-15

2.x=4x-12

3.0=3x-12

4.12=3x

5.x=4

The answer is 4

🐺🐺🐺

0.9=9/10.

I think can be like this:

$0.999...=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$

$0.999...=​​\frac{9}{10}+(\frac{9}{10})(\frac{1}{10})+(\frac{9}{10})(\frac{1}{10})^2+...$

Using infinite geometric series, with a=$\frac{9}{10}$ and r=$\frac{1}{10}$:

$0.999...=\frac{\frac{9}{10}}{(1-\frac{1}{10})}=\frac{9}{10}\times\frac{10}{9}=1$

Hope this help.

Why would anyone want to help you?

You have asked 57 questions and never EVER said thank you for an answer.

Plus. although you have given yourself some points for asking your own questions, 

You have never given anyone a point of answering.

I think that jugoslav's answer is correct. 

That circular top is included in the top surface of the box (area under the cylinder).

We can say $b\%$ of the master mix is Blue mix, and $r\%$ of the mastermix is Rye mix, so we have

$b+r=100$

$\frac{11b}{20}+\frac{3r}{10}=45$

we can multiply both sides of the second equation by 20 so that we are dealing with whole numbers instead of fractions, so we have:

$11b+6r=900$

We can multiply both sides of the first equation by 6, so that we can eliminate the r:

$6b+6r=600$

$11b+6r-6b-6r=900-600=5b=300$

$b\%=60\%$

so, out of each kg of mastermix or every 1000 grams of mastermix, there are $1000\cdot\frac{60}{100}=\boxed{600g}$ or $\boxed{C}$

Given a right triangle ABC with altitude BD drawn to hypotenuse AC. If AC = 11and DC = 4, what is the length of BC in simplest radical form? 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AC = 11         DC = 4        BC => x

x/11 = 4/x

x² = 44       x = √44        

BC = 2√11

Hello AngleObtuse!

I'm also learning about similar shapes in class. :))

The reason having the angles the same doesn't translate over to quadrilaterals is because a square and a non square rectangle would have all 90 degree angles, but not similar side lengths. 

I think the only way to prove that they're similar is to check that corresponding angles are congruent and corresponding sides are proportional.

I hope this helped. 

=^._.^=

We can write $\sqrt[3]{3}$ as $a$, and $\sqrt[3]{2}$ as $b$, and we know that

$(a-b)(a^2+b^2+ab)=a^3-b^3$, so

$a-b=\frac{a^3-b^3}{a^2+b^2+ab}=\sqrt[3]{3}-\sqrt[3]{2}=\frac1{\sqrt[3]{9}+\sqrt[3]{4}+\sqrt[3]{6}}$

that's only the denominator, so

$\frac1{\sqrt[3]{3}-\sqrt[3]{2}}=\frac1{\frac1{\sqrt[3]{9}+\sqrt[3]{4}+\sqrt[3]{6}}}=\frac{\sqrt[3]{\boxed{9}}+\sqrt[3]{\boxed{4}}+\sqrt[3]{\boxed{6}}}{\boxed1}$

so

A+B+C+D=9+4+6+1=$\boxed{20}$

If the number increased by 9, then x+9>28.

The midpoint is 2 right and 4 down from A, so B is 4 right 8 down from a=$\boxed{(-2,-4)}$

 a + bi + c + di + e + fi = 3i

(a+c+e) + (b+d+f)i = 3i

(a+c+-a - c) + (b+d+f)i = 3i

(b+d+f)i = 3i

b+d+f = 3

2+d+f = 3

d+f = 1

I hope this helped. :))

=^._.^=

Grey area = 3/4(2 * r²pi)       r = 2

∠KOP = 50º           ∠OKP = ∠OPK = 65º            ∠OQP = ∠OPQ = 110 - 65 = 45º

∠POQ = 90º

30x = 90       x = 3

w(2w + 1) = 36

2w² + w = 36

2w² + w - 36 = 0

w = 4