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Igoryamba
3 years ago
15

What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C

5(2)5(3)
Mathematics
2 answers:
butalik [34]3 years ago
5 0

ANSWER

10C_5(2)^{5}( - 3)^5

EXPLANATION

The given binomial expansion is:

{(2x - 3y)}^{10}

Compare this to

{(a + b)}^{n}

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

{x}^{5}  {y}^{5}

This implies that, r=5.

The terms in the expansion can be obtained using

T_{r+1}=nC_ra^{n-r}b^r

We substitute the given values to obtain;

T_{5+1}=10C_5(2x)^{10-5}(  - 3y)^5

T_{6}=10C_5(2x)^{5}(3y)^5

T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5}  {y}^{5}

Hence the coefficient is;

10C_5(2)^{5}( - 3)^5

kogti [31]3 years ago
3 0

Answer:

B

Step-by-step explanation:

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Answer:

H = h + x

h = 568 meters

Step-by-step explanation:

The helicopter descends 114 meters to be 454 meters above the ground.

Therefore, the equation which describes this situation using h as the original height will be given by  

H = h + x ....... (1)

Where H is the final height from the ground and x is the height that the helicopter ascends(Positive sign) or descends(Negative sign). (Answer)

Here, in our case, H = 454 meters and x = -114 meters.

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3 years ago
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Andrej [43]
We have that
<span>(c-4)/(c-2)=(c-2)/(c+2) - 1/(2-c)
 </span>- 1/(2-c)=-1/-(c-2)=1/(c-2)

(c-4)/(c-2)=(c-2)/(c+2)+ 1/(c-2)------- > (c-4)/(c-2)-1/(c-2)=(c-2)/(c+2) 
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(c-5)/(c-2)=(c-2)/(c+2)------------- > remember (before simplifying) for the solution that c can not be 2 or -2
(c-5)*(c+2)=(c-2)*(c-2)------------------ > c²+2c-5c-10=c²-4c+4
-3c-10=-4c+4----------------------------- > -3c+4c=4+10----------- > c=14

the solution is c=14
 
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3 years ago
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Find the roots of the equation x^3-3x^2+x+5
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x^2-4x+5=0, 
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x-2=i or -1,
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3 years ago
What is equivalent to 4 2/3
Svet_ta [14]
8
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3 years ago
Scenario: The Cannon Instructions: View the video found on page 1 of this Journal activity. Using the information provided in th
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Answer:

Part A

Yes

Part B

He will land at a point 30 meters from Cannon

Step-by-step explanation:

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The given function that represent the path of the Cannon, which is the path of a parabolic arc, is presented as follows;

f(x) = -0.05·(x² - 26·x - 120)

We note that the height, 'h', of the path of a parabola at a horizontal distance, 'x' from the origin is equal to f(x)

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Part B

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