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2 years ago

How many positive integers, including 1, are divisors of both 40 and 72?

Mathematics

1 answer:

Nezavi [6.7K]2 years ago

8 0

Answer:

1,2,4,8

Step-by-step explanation:

1 x 40 = 40

1 x 72 = 72

2 x 20 = 40

2 x 36 = 72

4 x 10 = 40

4 x 18 = 72

8 x 5 = 40

8 x 9 = 72

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3 years ago

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1 year ago

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A family travels 989.5 miles in 17.8 hours. Estimate the number of miles they can travel in one hour.

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2 years ago

Find 2 common angles that sum to (17pi/12) 2. Evaluate tan(17pi/12) using the sum identity for tangent.

Dmitry [639]

Note that
$\frac{17 \pi }{12} = \frac{3 \pi }{12} + \frac{14 \pi }{12} = \frac{ \pi }{4} + \frac{7 \pi }{6}$

Note that
$x= \frac{ \pi }{4}:\,\, sin(x) =cos(x)= \frac{1}{ \sqrt{2} },\,tan(x)=1\\x= \frac{7 \pi }{6} :\,\,sin(x)=- \frac{1}{2} ,\,\,cos(x)=- \frac{ \sqrt{3} }{2} ,\,\,tan(x)= \frac{1}{ \sqrt{3} }$

Use the identity
$tan(x+y)= \frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

Therefore
$tan( \frac{17 \pi }{12} )= \frac{1+ \frac{1}{ \sqrt{3} } }{1- \frac{1}{ \sqrt{3} } } = \frac{ \sqrt{3}+1 }{ \sqrt{3}-1} = \frac{( \sqrt{3}+1 )^{2}}{( \sqrt{3}-1 )( \sqrt{3}+1 )} = \frac{3+1+2 \sqrt{3}}{3-1} =2+ \sqrt{3}$

Answer: 2 + √3

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago