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2 years ago

What is the value of y? Please enter your answer, as an exact value

Mathematics

2 answers:

Natali [406]2 years ago

5 0

Answer: 8.66

So to Solve this we are going to be using Tangent.

Tangent: opposite/adjacent

Tan60 = y/5

Tan60 * 5 = y/5 *5

Tan60 * 5 = y

8.66 = y

Hatshy [7]2 years ago

4 0

Tan (60) = y/ 5
y = tan(60) * 5
and you know tan(60) = √3
so
y = 5√3

answer
5√3 (exact value)

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Find the exact value by using a half-angle identity.<br><br> tan seven pi divided by eight

murzikaleks [220]

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Answer:

1 -√2

Step-by-step explanation:

$\tan(x/2)=\dfrac{1-\cos(x)}{\sin(x)}\\\\\tan\left(\dfrac{1}{2}\cdot\dfrac{7\pi}{4}\right)=\dfrac{1-\cos\dfrac{7\pi}{4}}{\sin\dfrac{7\pi}{4}}=\dfrac{1-\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}}=\boxed{1-\sqrt{2}}$

tan(7π/8) = 1 -√2

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How could you use a number line to show that -4 + 3 and -3 + (-4)

Nonamiya [84]

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3 years ago

Un abuelo quiere repartir una cierta cantidad de manera proporcional entre sus tres nietos de 14, 10 y 6 años de edad ¿A quién l

stira [4]

Answer:

El nieto de 14 años de edad recibirá la mayor cantidad.

Step-by-step explanation:

En este problema, debemos tener en mente que la cantidad aportada es directamente proporcional a la edad de quien recibe, esta cantidad es igual al producto de la cantidad total y la razón de la edad de quien recibe y la suma de todas las edades.

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3 years ago

The shortest side of an isosceles triangle is 2x - 3 inches long. the longer sides are 6 inches longer than the shortest side. t

Zolol [24]

Side 1 = short side = 2x-3
side 2 = longer side = (short side) + 6 = (2x-3)+6 = 2x+3
side 3 = side 2 = 2x+3

Side 2 and side 3 are the longer two congruent sides

Add up the three sides and set them equal to the given perimeter of 33. Solve for x

(side1)+(side2)+(side3) = perimeter
(2x-3)+(2x+3)+(2x+3) = 33
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6x+3 = 33
6x+3-3 = 33-3
6x = 30
6x/6 = 30/6
x = 5

If x = 5, then the longer sides are 2*x+3 = 2*5+3 = 10+3 = 13 inches each
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3 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

2 years ago