[tex]\frac{4}{\sqrt{5}-1 } -\frac{4}{\sqrt{5}+1\\ }[/tex]

Answer:

I can't help you if there's no picture lol

I must speak in your language

ddddwfdwfd qdqfqfqqdqdqdqddqdqdq

Answer:la cantidad de invitados y la cantidad de mesas que no nos dio por favor agréguelo para que pueda ayudarlo :))

Step-by-step explanation:

Answer:

Step-by-step explanation:

Gogli complexes packages proteins, while endoplasmic reticulums transport proteins made by the gogli complexes around the cell.

The probability distribution for all possible values of X are given in the discrete probability table ;

Fair coin = {H, T}

card = {J, Q, K, A}

Sample space :

Probability of winning ; P(Winning) = 2/8 = 1/4 = 0.25

Probability of not winning ;

1 - P(wininng) = 1 - 1/4 = 3/4 = 0.75

X : ____ winning ____ not winning

P(X) : ____ 0.25 ________ 0.75

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Answer:

Step-by-step explanation:

65% of 51 = [tex]\dfrac{65}{100}*51[/tex]

                = 0.65 *51

                = 33.15

Answer:

y - 4 = -5(x + 1)

Step-by-step explanation:

Slope = m =  -5

(-1, 4)

y - y1 = m(x - x1)

y - 4 = -5(x - [-1])

y - 4 = -5(x + 1)

Answer:

18+3

Step-by-step explanation:

3x6= 18

and 18+3=21

Answer:

Answer:18.5

Step-by-step explanation:

10+8=18

18*5=90

90/4

22.5-4=18.5

Have a happy day! <3Miss Hawaii

Answer:

16

Step-by-step explanation:

sum of 8 and 10=18

5 times of sum of 8 and

10= 18×5=80

1/4 of 5 times of sum of 8 and

10= 1/4×80=20

and then subtracted 4 from

20 =20-4=16

Answer:

x = 5

Step-by-step explanation:

[tex]6x - 20 = 2x \\ - 20 = 2x - 6x \\ - 20 = - 4x \\ \frac{ - 20}{ - 4} = x = 5[/tex]

Step-by-step explanation:

......

Answer: C. 85°

Step-by-step explanation:

Answer:

I look lik dis almost dreads and then them types of braces and my eye brows look like that and I got a tattoo on my face but my eyes are blue and I have contacts so no more glasses

Answer:

60 cents per hour

Step-by-step explanation:

final price minus starting price is 3 divide by 5 hours to get .60

Answer:

k=4+h

...................

Answer:

After turn one she will have 0 points !

Step-by-step explanation:

After turn one she will have 0 points

After turn 2 she will have 8 points

After turn 3 she will have 7 points

She will not earn bonus points because she could not reach ten points

(Hope this helps can I pls have brainlist (crown)☺️)

Answer:

1

Step-by-step explanation:

use BEDMAS here:

4 divided by 8 is 1/2 or 0.5

2 divided by 4 is 1/2 or 0.5

now adding these two gives you:

1/2 + 1/2 = 1

Answer: 20

Step-by-step explanation:

Answer:

x=14

y=41

x+y= 55

Step-by-step explanation:

Since a and b are parallel you know that 123° and 3y are the same degree, so 123 divided by 3 because its 3×y= 41 so y=41

the line123° and 4x+1 are on has to equal 180 because its a straight line, 180-123= 57. 57- 1 because its not with the x= 56. 56÷4 because its 4×x/its with the x= 14. Hope thsi helps :)

Answer:

AAS

Step-by-step explanation:

I'm pretty sure its this, but im sorry if im wrong!

Since only one box contains gold, then the gold must be in Box 2.

Let's analyze the statements in each box:

Box 1 says, "The gold is not in Box 2." If this statement is true, it means the gold cannot be in Box 2.

Box 2 says, "The gold is in this box." If this statement is true, it means the gold must be in Box 2.

Box 3 says, "The gold is not in this box." If this statement is true, it means the gold cannot be in Box 3.

Now, let's consider the possibilities:

If Box 1 is true, then the gold is not in Box 2 (as stated in Box 1), and it is also not in Box 3 (since Box 3 would be false, and only one box contains the gold).

As per Box's 1 statement, the gold is not in Box 2.

If Box 3 is true, then the gold is not in Box 3 (as stated in Box 3), and it is also not in Box 2 (since Box 2 would be false, and only one box contains the gold).

Therefore, the gold must be in Box 1.

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#SPJ3

Answer:

Either [tex]x = (-\pi / 6)[/tex] or [tex]x = (\pi / 6)[/tex].

Step-by-step explanation:

Angle sum identity for cosine:

[tex]\cos(a + b) = \cos(a)\, \cos(b) - \sin(a) \, \sin(b)[/tex].

Double-angle identity for cosine and sine:

[tex]\begin{aligned}& \cos(2\, a)\\ &= \cos^{2}(a) - \sin^{2} (a) \\ &= (\cos^{2}(a) + \sin^{2}(a) - 2\, \sin^{2}(a) \\ &= 1 - 2\, \sin^{2}(a)\end{aligned}[/tex].

[tex]\sin(2\, a) = 2\, \cos(a) \, \sin(a)[/tex].

The secant of an angle is the inverse reciprocal of the cosine of that angle:

[tex]\displaystyle \text{$\sec(x) = \frac{1}{\cos(x)}$ if $\cos(x) \ne 0$}[/tex].

Apply these identities to rewrite the left-hand side of the equation.

[tex]\begin{aligned} & \sec(x) \, \cos(3\, x) \\ =\; & \frac{1}{\cos(x)} \cdot \cos(3\, x) \\ =\; & \frac{1}{\cos(x)} \cdot (\cos(x)\, \cos(2\, x) - \sin(x)\, \sin(2\, x)) \\ =\; &\cos(2\, x) \\ & - \frac{1}{\cos(x)} \cdot (\sin(x)\, \sin(2\, x)) \\ =\; & \cos(2\, x)\\ & - \frac{1}{\cos(x)} \cdot (\sin(x) \cdot (2\, \sin(x)\, \cos(x))) \\ =\; & \cos(2\, x) - 2\, \sin^{2}(x) \\ =\; & (1 - 2\, \sin^{2}(x)) - 2\, \sin^{2}(x) \\ =\; &1 - 4\, \sin^{2}(x) \end{aligned}[/tex].

Solve the rewritten equation:

[tex]\displaystyle \sin^{2}(x) = \frac{1}{4}[/tex].

Either [tex]\sin(x) = (1/2)[/tex] or [tex]\sin(x) = (-1/2)[/tex].

The question required that [tex](-\pi / 2) \le x \le (\pi / 2)[/tex]. The sine function is one-to-one over this range. In other words, for each [tex]y[/tex] between [tex](-1)[/tex] and [tex]1[/tex], there is exactly one [tex]x[/tex] between [tex](-\pi / 2)[/tex] and [tex](\pi / 2)[/tex] such that [tex]\sin(x) = y[/tex].

If [tex]\sin(x) = (-1/2)[/tex] and [tex](-\pi / 2) \le x \le (\pi / 2)[/tex], then [tex]x = (-\pi / 6)[/tex].

Otherwise, if [tex]\sin(x) = (1/2)[/tex] and [tex](-\pi / 2) \le x \le (\pi / 2)[/tex], then [tex]x = (\pi / 6)[/tex],

Verify that [tex]\cos(x) \ne 0[/tex] for either of the two [tex]x[/tex] values, such that the [tex]\sec(x)[/tex] in the original equation takes a real value.

Answer:

Sale kute bhan के lole

Tera baap aya ajju(^bhai 94

Answer:

f = 3

Step-by-step explanation:

7f + 6 = 27 ( subtract 6 from both sides )

7f = 21 ( divide both sides by 7 )

f = 3

7f+6=27

7f=27-6

7f=21

Divide both sides by 7 to isolate f:

f=3

Hope this helps!

#HaveAnAmazingDay

[tex]GraceRosalia[/tex]

Answer:

60

Step-by-step explanation:

S₄ = 4(4+1)(4+2) = 120

S₃ = 3(3+1)(3+2) = 60

4th term = 120-60=60

Answer:2x-1

Step-by-step explanation:

Answer:

1.5 Or C :)

Step-by-step explanation:

Hello Carousel!

[tex] \huge \boxed{\mathbb{QUESTION} \downarrow}[/tex]

[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]

[tex] \tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex]

First, let's take I as ⇨ [tex]\tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex].

[tex]\tt \:I = \int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x +1}} \: dx \\\tt \: I =\int \: \frac{ x + 2 }{( {x}^{2} + 2x + 1 + x + 2) \sqrt{x + 1}} \: dx \\ \tt \:I =\int \: \frac{ x + 2 }{( ({x + 1}^{2}) + x + 2) \sqrt{x + 1}}[/tex]

Let, x + 1 = m² => dx = 2mdm.

[tex]\tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: m} 2mdm \\ \tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: \bcancel{ m}} 2 \bcancel{m}dm \\ \tt \: I = \: \int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 } 2dm[/tex]

Now, divide the numerator & denominator by m²....we'll get it as...

[tex]\tt \:I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{ {m}^{2} + 1 + \frac{1}{ {m}^{2} } } \: dm \\\tt \: I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{( {m}^{2} + \frac{1}{ {m}^{2} } - 2) + 3} \: dm \\\tt \: I =2\int \: \frac{ (1 + \frac{1}{ {m}^{2} }) \: dm }{ ({m} - \frac{1}{ m } ) ^{2} + 3} [/tex]

Now, let m - 1/m be t => (1 + 1/m²) dm = dt

[tex]\tt \:I =2\int \: \frac{ dt}{ {t}^{2} + 3 } \\ \tt \:I = 2\int \: \frac{ dt}{ {t}^{2} + ( \sqrt{3}) ^{2} }[/tex]

We know, [tex]\tt \:\int \: \frac{dx}{ {x}^{2} + a ^{2} } = \frac{1}{a} tan ^{ - 1} (\frac{x}{a} ) + c[/tex]...therefore...

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{t}{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 1}[/tex]

Now, substitute the value of 't' in eq. 1..we'll get..

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{m - \frac{1}{m} }{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 2}[/tex]

Now, substitute the value of 'm' in eq. 2...we'll get...

[tex] \tt \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{ \sqrt{x + 1} - \frac{1}{ \sqrt{x - 1} } }{ \sqrt{3} } ) + c \\ \tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x + 1 - 1}{ \sqrt{3} \sqrt{x - 1} } ) + c \\ \boxed{\boxed{ \bf \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x }{ \sqrt{3 (x - 1} )} ) + c }}[/tex]

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Hope it'll help you!

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