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kenny6666 [7]
3 years ago
13

Determine the Taylor series to represent cos(pi/3+h)

Mathematics
1 answer:
Dmitrij [34]3 years ago
7 0
Develop <span> cos(</span>π<span>/3 + h) =

=  1/2 -h.</span>√3/2 - h²/4√3 +h³/4√3 + h⁴/48 - h⁵/80√3 - h⁶/1440 +....
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6
GalinKa [24]

Answer:

I don't know sorry

sorry

Step-by-step explanation:

I don't know sorry

sorry

3 0
3 years ago
If(√14/√7-2)-(√14/√7+2)=a√7+b√2 find the values of a and b where a and b are rational numbers​
seraphim [82]

Answer:

  • a = 4/3 and b = 0

============================

<h2>Given expression:</h2>

\dfrac{\sqrt{14} }{\sqrt{7}-2} -\dfrac{\sqrt{14} }{\sqrt{7}+2}

<h2>Simplify it in steps:</h2>

<h3>Step 1</h3>

Bring both fractions into common denominator:

\dfrac{\sqrt{14} (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} - \dfrac{\sqrt{14} (\sqrt{7}-2)}{(\sqrt{7}-2)(\sqrt{7}+2)}

<h3>Step 2</h3>

Simplify:

\dfrac{\sqrt{14} ((\sqrt{7}+2) - (\sqrt{7}-2))}{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{\sqrt{14} (\sqrt{7}+2 - \sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{4\sqrt{14} }{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{4\sqrt{14} }{(\sqrt{7})^2-2^2} =

\dfrac{4\sqrt{14} }{7-4} =

\dfrac{4}{3}  \sqrt{14} }

<h3>Step 3</h3>

Compare the result with given expression to get:

  • a = 4/3 and b = 0

4 0
2 years ago
Pls6ssssssssssssssssssssssssssss
max2010maxim [7]

Answer: 220

Step-by-step explanation:

176 dived by 4 is 44, 44 times 5 is 220 your welcome

8 0
3 years ago
Read 2 more answers
In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find the length of the angle bisector of an
nordsb [41]

Check the picture below.

bearing in mind that an angle bisector segment will cut the angle A in two equal halves, thus θ will be half of A.

6 0
4 years ago
Could someone help me please ? :)
Degger [83]

ANSWER = 17.5% of 360 = 63

EXPLANATION:

Find 10% of 360 = 36

Then 5% of 360 = 18

Then 2.5% of 360 = 9

Add them together

so 17.5% of 360 = 63

8 0
3 years ago
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