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iVinArrow [24]
3 years ago
7

A triangle has side lengths of 11 cm, 60 cm, and 61 cm. which is the correct equation to verify whether the side lengths form a

right triangle
PLEASE HELP

Mathematics
1 answer:
inysia [295]3 years ago
4 0
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Pythagoras' theorem:

a^2 + b^2 = c^2

'c' is the hypotenuse and also the longest length

Taking this into consideration, the correct option would be:

C) 11^2 + 60^2 = 61^2

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Answer:

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Step-by-step explanation:

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2 years ago
If it takes 8 inches of ribbon to make a bow,how many bows can be made from 3 feet of ribbon (1 foot = 12 inches)? Will any ribb
Aloiza [94]

First, find out how much ribbon there is.

3 x 12 = 36

Now, see how many times you can divde that by 8.

36 can only be divided by 8 four times, and there would be 4 inches of ribbon left.

So, you would be able to make 4 bows, with 4 inches of ribbon remaining.

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2 years ago
Help me solve this algebra 1 question please
s344n2d4d5 [400]
You should know that simply your answer
3 0
2 years ago
xYou volunteer to help drive children at a charity event to the​ zoo, but you can fit only 6 of the 16 children present in your
brilliants [131]

There are 8008 groups in total, in other to drive the children

<h3>How to determine the number of groups?</h3>

From the question, we have

  • Total number of children, n = 16
  • Numbers to children at once, r = 6

The number of group of children that could be carried at once is calculated using the following combination formula

Total = ⁿCᵣ

Where

n = 16 and r = 6

Substitute the known values in the above equation

Total = ¹⁶C₆

Apply the combination formula

ⁿCᵣ = n!/(n - r)!r!

So, we have

Total = 16!/10!6!

Evaluate

Total = 8008

Hence, the number of groups is 8008

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8 0
11 months ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

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3 0
1 year ago
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