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laila [671]
3 years ago
10

Plzz help me to find coefficient in binomial theorem (25 points)

Mathematics
1 answer:
Viefleur [7K]3 years ago
5 0

<em>Denote x2 by y. </em>

<em>  </em>

<em>(x2-3)7=(y-3)7 </em>

<em>  </em>

<em>This is a binomial expansion in y, and you want the coefficient of y4 because y4=x8 </em>

<em>  </em>

<em>You have 7 terms of (y-3) in (y-3)7.  To get the fourth power of y, you need to choose y from four of the terms.  The number of ways you can do this is the combinations of 7 things taken 4 at a time.  This is: </em>

<em>  </em>

<em>7!/(4!3!)=35</em>

<em />

<em>So, the coefficient of x8 in the given expansion will be 210.</em>

<em>HOPE IT HELPS</em>

<em>THANK YOU </em>

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B.
Helen [10]

Answer:

1. ΔUJN ≅ ΔUGN by SAS rule of congruency

2. Side Angle Side Rule of congruency

3. JN ≅ GN, ∠JUN ≅ ∠NUG, ∠UNJ ≅ ∠UNG, NU ≅ NU

4. UN bisects angle ∠JUG and UN bisects segment JG

Step-by-step explanation:

The given figure consists of three triangles

1. ΔUJN, 2. ΔUGN, and 3. ΔUJG

Whereby segment UN is perpendicular to segment JG, we have;

ΔUJN and ΔUGN are right triangles

∴ ∠UNJ = ∠UNG = 90° by definition of segment UN being perpendicular to segment JG

NU ≅ NU by reflexive property

Whereby UN bisects ∠JUG and segment JG, we have;

JN = GN by definition of a bisected segment

Therefore, we have;

1. ΔUJN ≅ ΔUGN by SAS rule of congruency

2. Side Angle Side Rule of congruency

3. JN ≅ GN, ∠JUN ≅ ∠NUG, ∠UNJ ≅ ∠UNG, NU ≅ NU

4. UN bisects angle ∠JUG and UN bisects segment JG

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2 years ago
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Answer:

D. 1/3

Step-by-step explanation:

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What is the snswer to 3a+2b x 7c
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I am pretty sure your answer is this because 3+2=5 and 5*7=35 abc
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3 years ago
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