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Natali [406]
3 years ago
6

[2/SEC(¶/3)•[lim x→0 x^3+8x+10]^2]/[lim θ→0 sinθ/θ]

Mathematics
1 answer:
Oliga [24]3 years ago
4 0
Assuming the pilcrow is supposed to represent pi:

\frac{\frac{2}{sec(\frac{\pi}{3})}*(\lim_{x \to 0} x^3+8x+10)^2}{ \lim_{\theta \to 0} \frac{sine(\theta)}{\theta}}\\\frac{\frac{2}{\frac{1}{cosine(\frac{\pi}{3})}}*((0)^3+8(0)+10)^2}{\frac{sine((0))}{(0)}}\\\frac{\frac{2}{\frac{1}{\frac{1}{2}}}*(0+0+10)^2}{\frac{0}{0}}\\\frac{\frac{2}{2}*(10)^2}{ \lim_{\theta \to 0} \frac{cosine(\theta)}{1}}\\\frac{100}{cosine((0))}\\\frac{100}{1}\\100

100
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QUESTION 10
adelina 88 [10]

Answer:

Step-by-step explanation:

A = L * W

A = 22

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now we sub

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3x = -7               2x = 3

x = -7/3              x = 3/2

I believe that x = -7/3 is an extraneous solution because when u plug it into ur length and width u get a negative number....and I dont think the sides of ur rectangle have negative values, however, it does work in ur equation and it equals 22 when multiplied...so I am not 100% sure

length = 2x + 1.......2(3/2) + 1........3 + 1 = 4

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6 0
3 years ago
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6 is greater than 5, so

5¾<6¼
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2 years ago
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