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tankabanditka [31]
3 years ago
15

If a 5 ft tall man cast an 8 ft long shadow at the same time a tree cast a 24 ft long shadow, how tall is the tree?

Mathematics
2 answers:
Taya2010 [7]3 years ago
5 0

Answer:

15 feet

Step-by-step explanation:

We have 2 similar right triangles with legs height and length of shadows.

height of men : length of shadows of the man = height of tree : length of shadows of the tree

5 : 8 = x : 24

8x = 5* 24

x = 5*24/8 = 15 (feet)

erma4kov [3.2K]3 years ago
3 0

Answer:

15ft

Step-by-step explanation:

5 ft  is to 8 ft

A ft is to  24 ft

A = 24*5/8

A = 15ft

15ft

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Find the area of a triangle <br> b = 30m<br> h = 15.6m
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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
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\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
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\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
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A cuboid is of dimensions 80 cm x 44 cm x 25 cm. The number of small
konstantin123 [22]

Answer:

Answer: 704

Step-by-step explanation:

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