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harkovskaia [24]
3 years ago
11

In ΔMNO, the measure of ∠O=90°, the measure of ∠M=46°, and NO = 96 feet. Find the length of OM to the nearest tenth of a foot.

Mathematics
1 answer:
GalinKa [24]3 years ago
6 0

Answer:

92.7 ft

Step-by-step explanation:

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Cole is studying employment trends.
natali 33 [55]

Answer:

225

Step-by-step explanation:

So we know that 20% of the social workers he surveyed was 45 people. The easiest way to find all 100% of the social workers is to do 45 x 5 because 20% x 5 = 100%. So 45 x 5 = 225.

Hope this helped! :)

3 0
3 years ago
What is 92.3 times 4 to the power of ten
steposvetlana [31]

I am pretty sure your answer will be 96783564.8, but it may need to be rounded. If I am wrong, I am sorry.


Hope this helps~!

5 0
3 years ago
Linear equations and interception
iVinArrow [24]

Answer:

x - intercept = 6

y - intercept = 3

Step-by-step explanation:

To find x - intercept, substitute y = 0 in the equation and solve for x.

x = 6

Similarly, To find y - intercept, substitute x = 0 in the equation and solve for y.

y = 3

5 0
2 years ago
I need answers for Question 1 and 2
Xelga [282]

Answer:  The answer to your first question would be A. 1.  This is simply because anything raised to the power of 0 is 1 no matter what.  The answer to your second question would be B. 1, for the same reason as above, anything to the power of 0 is 1.  Hope this helped and make sure to remember that rule, it is important and easy to remember!

4 0
3 years 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

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
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