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Volgvan
3 years ago
15

I need help so I can finish my algebra hw

Mathematics
1 answer:
Sloan [31]3 years ago
7 0
None of those are correct. the answer would be 3 x^{2} -24x+36=180.

you can draw a picture of the garden space, where x is the width of the space, 3x is the length. to account for the 3 inch border you need to subtract 3 from both ends, so the width of his garden is then x-6, and the length is 3x-6.

multiplying these together, which is the formula for area, gets you 3 x^{2} -24x+36=180. 

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hram777 [196]
The graph of f(x) translates down 4 units to give f(x) - 4
So the x coordinate stays the same but the y coordinate is 4 units less

So the corresponding point is (-3, 1)
7 0
3 years ago
Find a cubic function with the given zeros. (1 point) Square root of seven. , - Square root of seven. , -4
Debora [2.8K]

Given :

Three roots , \sqrt{7},\ -\sqrt{7},\ -4 .

To Find :

A cubic function with the given zeros.

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Equation of polynomial of 3 zeroes is given by :

(x-a_1)(x-a_2)(x-a_3)=0\\\\(x-\sqrt{7})(x+\sqrt{7})(x+4)=0\\\\( x^2-7)(x+4)=0\\\\x^3+4x^2-7x-28=0

Therefore, the cubic function of given zeroes is x^3+4x^2-7x-28=0.

Hence, this is the required solution.

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4 years ago
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Answer:

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3 years ago
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Answer:

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7 0
3 years ago
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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;-------------------------------\\\\

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\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
3 years ago
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