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alexdok [17]
4 years ago
8

(Please help!) 11(5+8)+17s

Mathematics
1 answer:
Svetradugi [14.3K]4 years ago
5 0

143 + 17s

I can't figure out <em>s </em>based on your question alone, so I'm going to assume that the two factors are to be separated.

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Simplify tan9x-tan5x / 1+tan9xtan5x<br> (SHOW WORK)
Charra [1.4K]

Answer:

The simplest form is tan(4x)

Step-by-step explanation:

* Lets revise the identity of the compound angles

- tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)tan(b)}

- tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}

* Lets solve the problem

- Let 9x = 5x + 4x

∴ tan(9x) = tan(5x + 4x)

- Use the rule of the compound angle

∵ \frac{tan(9x)-tan(5x)}{1+tan(9x)tan(5x)} ⇒ (1)

∵ tan(5x+4x)=\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)} ⇒ (2)

∵ tan(9x) = equation (2)

- Substitute (2) in (1)

∴ \frac{\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)}-tan(5x)}{1+(\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)})tan(5x)}

- Multiply up and down by (1 - tan(5x)tan(4x))

∴ \frac{tan(5x)+tan(4x)-tan(5x)[1-tan(5x)tan(4x)]}{1-tan(5x)tan(4x)+tan(5x)[tan(5x)+tan(4x)]}

- Simplify up and down

∴ \frac{tan(5x)+tan(4x)-tan(5x)+tan^{2}(5x)tan(4x)}{1-tan(5x)tan(4x)+tan^{2}(5x)+tan(5x)tan(4x) }

∴ \frac{tan(4x)+tan^{2}(5x)tan(4x)}{[1+tan^{2}(5x)]}

- Take tan(4x) as a common factor up

∴ \frac{tan(4x)[1+tan^{2}(5x)]}{[1+tan^{2}(5x)]}

- Cancel [1 + tan²(5x)] up and down

∴ The answer is tan(4x)

4 0
4 years ago
Reuben bought n packs of pencils, each pack had 15 pencils, write an equesion to represent the total number of penciles p that r
zimovet [89]
P=15n. Hope that helps :)
5 0
4 years ago
Large balloons are sold in packages of 12. Select the expressions that can represent the total number of balloons in p packages
WARRIOR [948]
B is the answer to your question
5 0
3 years ago
"We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 7/8 of its previous height at each
tatyana61 [14]

Answer:

Total Time = 4.51 s

Step-by-step explanation:

Solution:

- It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics:

                             y = v_o*t + 0.5*a*t^2

Where,   v_o : Initial velocity = 0 ... dropped

              a: Acceleration due to gravity = 32 ft / s^2

              y = h ( Initial height )

                             h = 0 + 0.5*32*t^2

                             t^2 = 2*h / 32

                             t = 0.25*√h   ...... Proven

- We know that ball rebounds back to 7/8 of its previous height h. So we will calculate times for each bounce:

1st : 0.25*\sqrt{15}\\\\2nd: 0.25*\sqrt{15} + 0.25*\sqrt{15*\frac{7}{8} } + 0.25*\sqrt{15*\frac{7}{8} } = 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} }\\\\3rd: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 2*0.25*\sqrt{15*(\frac{7}{8} })^2\\\\= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2\\\\4th: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 2*0.25*\sqrt{15*(\frac{7}{8} })^3 \\\\

= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 0.5*\sqrt{15*(\frac{7}{8} })^3

- How long it has been bouncing at nth bounce, we will look at the pattern between 1st, 2nd and 3rd and 4th bounce times calculated above. We see it follows a geometric series with formula:

  Total Time ( nth bounce ) = Sum to nth ( \frac{1}{2}*\sqrt{15*(\frac{7}{8})^(^i^-^1^) }  - \frac{1}{4}*\sqrt{15})

- The formula for sum to infinity for geometric progression is:

                                   S∞ = a / 1 - r

Where, a = 15 , r = ( 7 / 8 )

                                   S∞ = 15 / 1 - (7/8) = 15 / (1/8)

                                   S∞ = 120

- Then we have:

                                  Total Time = 0.5*√S∞ - 0.25*√15

                                  Total Time = 0.5*√120 - 0.25*√15

                                  Total Time = 4.51 s

5 0
4 years ago
Write the equation in standard form.<br> y-4 = –3/5 (x + 10)
wlad13 [49]

Answer:

-4y+10x= -3/5

Step-by-step explanation:

Hopeful this helps

6 0
3 years ago
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