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Digiron [165]
3 years ago
14

Halppppp, thanks also no work no credit

Mathematics
1 answer:
rosijanka [135]3 years ago
7 0
I’m sorry but how r we supposed to answer there’s no picture
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the perimeter of a rectangle is at most 66 inches. the length is 2x + 6 inches and the width is x - 3 inches. write and solve an
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3 years ago
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rosijanka [135]
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---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\&#10;-------------------------------\\\\

\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[  \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]&#10;\\\\\\&#10;2\left[  \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]&#10;\\\\\\

\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]&#10;\\\\\\&#10;2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}
4 0
2 years ago
You walk into a store that is closing. the pants normally cost $60.00 but it is discounted 75% they are also taking additional 2
eimsori [14]

Answer:

The final cost of the pants is $ 12.

Step-by-step explanation:

Given that at a store that is closing the pants normally cost $ 60.00 but it is discounted 75%, and they are also taking an additional 20% off, to determine what is the final price for the pants, the following calculation must be performed:

100 - 20 = 80

(60 - (60 x 0.75)) x 0.80 = X

(60 - 45) x 0.80 = X

15 x 0.80 = X

12 = X

Thus, the final cost of the pants is $ 12.

7 0
2 years ago
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