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1 year ago

Sin(x+ pi/6)- sin(x-pi/6)= 1/2

Mathematics

1 answer:

agasfer [191]1 year ago

5 0

Answer:

You can download the answer here. Link below!

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Have a wonderful day and here are free pts to make your day better <3 12 x 7 = ?

Rom4ik [11]

Answer:

its 84 but tysm thats so sweet

Step-by-step explanation:

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2 years ago

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What is the range of the function f(x)=(x+4)²+5

hammer [34]

Answer:

$Range=\left \{y|y\geq 5\right \}$

Step-by-step explanation:

Since given the expression, this represents graphically a parabola with arms pointing up, and a vertical translation of its vertex 5 units up, then the range of the function must be all the y values such that they are greater or equal to 5. This is written as:

$Range=\left \{y|y\geq 5\right \}$

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2 years ago

PLEASE HELPPP!

Liono4ka [1.6K]

Answer:

Bottle B has a larger volume of lotion therefore the correct answer

Step-by-step explanation:

1×2×6=12/3=4 Bottle A

1.5×3×4=18/3=6 Bottle B

8 0

2 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago

If x€{-2,-1,0,1,2,3,4,5} find the solution set 7-x>0 ii 3-4x>-2 vi 3x+4>-2

maw [93]

Answer:

i) {-2 , -1 , 0 , 1 ,2 , 3 , 4 , 5}

ii) {-2 , -1 , 0 , 1}

iii) {-1 , 0 , 1 ,2 , 3 , 4 , 5}

Step-by-step explanation:

i) 7 - x > 0

⇔ 7 > x

⇔ x < 7

⇔ x ∈ {-2 , -1 , 0 , 1 ,2 , 3 , 4 , 5}

____________________________

ii) 3 - 4x > -2

⇔ 3 + 2 > 4x

⇔ x < 5/4

⇔ x ∈ {-2 , -1 , 0 , 1}

____________________________

iii) 3x + 4 > -2

⇔ 3x > -2 - 4

⇔ 3x > -6

⇔ x > -6/3

⇔ x > -2

⇔ x ∈ {-1 , 0 , 1 ,2 , 3 , 4 , 5}

8 0

1 year ago