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ycow [4]
3 years ago
10

The slopes of perpendicular line segments are 1/3 and d/5. What is the value of d

Mathematics
1 answer:
a_sh-v [17]3 years ago
4 0
If 2 lines (or line segments) are perpendicular, then the product of their slopes is equal to -1.

Thus, since the given line segments are perpendicular, we have:

             \displaystyle{  \frac{1}{3}\cdot  \frac{d}{5}=-1.

Multiplying both sides of the equation by 15, we get d=-15.


Answer: d=-15
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Answer:

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

Step-by-step explanation:

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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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