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4 years ago

5

A lucky integer is a positive integer which is divisible by the sum of its digits. what is the least positive multiple of 9 that

is not a lucky integer?

1 answer:

Art [367]4 years ago

5 0

99.

It is not divisible by 18. Before that they are all added up to 9.

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Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==> sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

sin(5<em>x</em>) = 0 <u>or</u> sin(2<em>x</em>) = 0 <u>or</u> 2 cos(2<em>x</em>) + 1 = 0

sin(5<em>x</em>) = 0 <u>or</u> sin(2<em>x</em>) = 0 <u>or</u> cos(2<em>x</em>) = -1/2

sin(5<em>x</em>) = 0 ==> 5<em>x</em> = arcsin(0) + 2<em>nπ</em> <u>or</u> 5<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … … ==> 5<em>x</em> = 2<em>nπ</em> <u>or</u> 5<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … … ==> <em>x</em> = 2<em>nπ</em>/5 <u>or</u> <em>x</em> = (2<em>n</em> + 1)<em>π</em>/5

sin(2<em>x</em>) = 0 ==> 2<em>x</em> = arcsin(0) + 2<em>nπ</em> <u>or</u> 2<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … … ==> 2<em>x</em> = 2<em>nπ</em> <u>or</u> 2<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … … ==> <em>x</em> = <em>nπ</em> <u>or</u> <em>x</em> = (2<em>n</em> + 1)<em>π</em>/2

cos(2<em>x</em>) = -1/2 ==> 2<em>x</em> = arccos(-1/2) + 2<em>nπ</em> <u>or</u> 2<em>x</em> = -arccos(-1/2) + 2<em>nπ</em>

… … … … … … ==> 2<em>x</em> = 2<em>π</em>/3 + 2<em>nπ</em> <u>or</u> 2<em>x</em> = -2<em>π</em>/3 + 2<em>nπ</em>

… … … … … … ==> <em>x</em> = <em>π</em>/3 + <em>nπ</em> <u>or</u> <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

5 0

3 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

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sergejj [24]

X=5
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Mamont248 [21]

The value of x is 21.

To find this one, we need to know the formula with intersecting lines outside of the circle.

It is that if you subtract the smaller arc from the larger arc and divide by 2, you will have the exterior angle.

Therefore, we can write and solve the following equation:

(5x - 63) / 2 =x
5x - 63 = 2x
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