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VashaNatasha [74]
3 years ago
5

A person who is 5 feet tall is standing 126 feet from the base of a​ tree, and the tree casts a 140 foot shadow. The​ person's s

hadow is 14 feet in length. What is the height of the​ tree?

Mathematics
1 answer:
Eva8 [605]3 years ago
8 0
Hope this helps but if you got any other questions feel free to ask :) cause I suckeeeeddd at Geometry but I had to pass lolllll

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Is every natural number a rational number.
kolezko [41]
True
False
False
True
False
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8 0
3 years ago
1. Iris has two ribbons. The first ribbon is 11.7 in. long. The second ribbon is 5.25 in. long. She wants to cut them so that th
lisov135 [29]
The estimated lengths are 12in. and 5 in. This would be an estimated difference of 7in.


The estimated difference is going to be higher than the actual difference because in rounding 11.7 to 12, you are losing .30in, but in rounding down 5.25 to 5, you are actually gaining .75in.


The ACTUAL difference is 6.45in
3 0
3 years ago
8. It takes Olivia 3 | yards of ribbon
Leokris [45]

Answer:

21 yards

Step-by-step explanation:

3 yards for 1 mum

3*7 yards for 7 mums

21 yards for 7 mums

I don't know if this is what you mean but here have this

7 0
3 years ago
Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that
Bond [772]

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

.

.

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

.

.

.

Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Find out more information about taylor series here

brainly.com/question/13057266

#SPJ4

3 0
2 years ago
What is -5 5/8 + 3 1/2
aksik [14]

Answer:

exact form: -17/8

decimal form: -2.125

mixed number form: -2 1/8

Step-by-step explanation:

cmon this is freaking addition

4 0
3 years ago
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