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tensa zangetsu [6.8K]
3 years ago
11

Y=3x+2 is the equation of a straight line graph. Where does it cross the y-axis?

Mathematics
1 answer:
tamaranim1 [39]3 years ago
8 0
The line crosses the y-axis at point (0,2). You know this because the slope intercept form is y=mx+b. y is any point on the y-axis. m is the slope of the line. x is and point on the x-axis. Finally, b is the y-intercept or where the line crosses the y-axis. As a result, your equation has a 2 for b. So, your answer would be: the line crosses the y-axis at point (0,2). 2 being where the line crosses the y-axis.
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AysviL [449]

You can take x = a, 2y = b and then can apply the binomial theorem.

The expansion of given expression is given by:

Option D: x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7 is

<h3>What is binomial theorem?</h3>

It provides algebraic expansion of exponentiated(integer) binomial.

According to binomial theorem,

(a+b)^n = \sum_{i=0}^n ^nC_i a^ib^{n-i}

<h3>How to use binomial theorem for given expression?</h3>

Taking a = x, and b =2y, we have n = 7, thus:

(x+2y)^7 = \: ^7C_0x^0(2y)^7 + \: ^7C_1x^1(2y)^6 + \: ^7C_2x^2(2y)^5 + \: ^7C_3x^3(2y)^4 + \:^7C_4x^4(2y)^3 + \:^7C_5x^5(2y)^2 + \:^7C_6x^6y^1 + \: ^7C_0x^7y^0\\\\&#10;(x+2y)^7 = 128y^7 + 448xy^6 + 672x^2y^5 + 560x^3y^4 + 280x^4y^3 + 84x^5y^2 + 14x^6y + x^7\\\\&#10;(x+2y)^7 = x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7

Thus, Option D: x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7 is correct.

Learn more about binomial theorem here:

brainly.com/question/86555

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Firstly , we need to draw diagram

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