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Alja [10]
3 years ago
15

Question is on attached photograph. ​

Mathematics
1 answer:
inysia [295]3 years ago
8 0

Answer:

Here you go sorry.

Step-by-step explanation:

Question 1.

y = 4x - 8 has a slope of 4. Since parallel lines have equal slopes, the slope of your new line will also be 4.

Now, you have a point (3,9) and a slope 4. Use the Point-Slope Form for the equation of a line.

y - y1 = m(x - x1)

y - 9 = 4(x - 3)

y - 9 = 4x - 12

y = 4x - 3

Question 2.

Since it is parallel, it must have the same slope.

y = (1/2)x + b

Find b from the given point.

4 = (1/2)(2) + b

b = 3

y = (1/2)x + 3

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Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

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