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Svetach [21]
3 years ago
6

Bamboo is one of the fastest-growing plants. A typical growth rate for bamboo in temperate climates is 3-10 centimeters per day

during the growth season. Which of the following equations, where t represents time in days, and L represents length in centimeters, could be descriptions of the growth of a bamboo plant? Select all that apply.  A.L=1.1t  B.L=2.5t  C.L=3.6t  D.L=7.1t  E.L=9.3t  F.L=10.4t
Mathematics
2 answers:
Maurinko [17]3 years ago
5 0
Since its 3 through 10 centimeters per day the answers would be

C.L=3.6t
D.L=7.1t
E.L=9.3t

(I am really not sure of this)
zepelin [54]3 years ago
3 0

Answers

L=7.1t

L=9.3t

Step-by-step explanation:


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2x+38=50
2x+38-38=50 -38
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svetoff [14.1K]
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3 years ago
If √2 = 1.414 then the value of 5+√2 / 5-√2 is
Gala2k [10]

Answer:

1.787

Step-by-step explanation:

\\  \sf \frac{5 +  \sqrt{2} }{5 -  \sqrt{2} }  \times  \frac{5 +  \sqrt{2} }{5  +   \sqrt{2} }

\\   \sf =  \frac{(5 +  \sqrt{2}) }{5 -  \sqrt{2} }  \times  \frac{(5 +  \sqrt{2}) }{5 +  \sqrt{2} }

\\  \sf =  \frac{25 + 5 \sqrt{2 } + 5 \sqrt{2} + 2  }{( {5})^{2}  - ( \sqrt{ {2}})^{2} }

\\  \sf =  \frac{27 + 10 \sqrt{2} }{25 - 2}

\\  \sf =  \frac{27 - 10 \sqrt{2} }{23}

Now putting the value of√2

\\  \sf =  \frac{27 + 10 \times 1.414}{23}

\\  \sf =  \frac{27 + 14.14}{23}

\\  \sf =  \frac{41.14}{23}

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3 0
2 years ago
876.34 rounded to the nearest thousandth
kogti [31]

Answer:

There is no thousands place. It only goes up to a hundred

Step-by-step explanation:

You need another number after 4 to be able to round thousands

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3 years ago
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likoan [24]

Answer:

x =5

Step-by-step explanation:

We can use proportions to solve this problem.  Put the side of one triangle over the same side of the other triangle.

x           4

------ = ----------

40           32

Using cross products

32x = 4*40

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Divide each side by 32

32x/32 =160/32

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3 years ago
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