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3 years ago

Solve for y. 3=7(1-1)(it isn't 2)

Mathematics

1 answer:

laiz [17]3 years ago

4 0

Answer:

First off, I can't solve for y, there is no y variable.

Second, 3 = 7 ( 1 - 1 ) is not true because 3 ≠ 0 ( 3 doesn't equal 0 ) :

3 = 7 ( 1 - 1 )

3 = 7 ( 0 ) ( Simplify Parenthesis )

3 = 0 ( Multiply 7 and 0 )

3 ≠ 0

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2 years ago

What is the GCF of (8x-6) <br><br>sos T-T

makvit [3.9K]

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the G.C.F. of 8x-6.

$\triangle~\fbox{\bf{KEY:}}$

G.C.F. stands for Greatest Common Factor.

So what is the G.C.F. of 8x-6? We can list all of their common factors and then select the greatest one, like so:

$\longmapsto\bf{Factors\;of\;8x:}$ 1, 2, 4, 8, x

$\longmapsto\bf{Factors\;of\;-6:}$ -1, 1, 2, -3, 3, -6, 6

So the GCF is:

Now, we can also factor it out by placing it outside the parentheses ():

2(4x-3)

See, we factored it out by dividing both terms of the expression by the G.C.F.

$\ddot\bigstar$ Remember this...

$\fbox{\bf{We\;factor\;out\;the\;G.C.F.\;by\;dividing\;all\;the\;terms\;of\;the\;expression\;by\;it}}$

Hope this helps you out! :D

Ask in comments if any queries arise.

~Just a smiley person helping fellow students :)

6 0

1 year ago

What is the binomial expansion of (x 2y)7? 2x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5 14xy6 2y7 x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5

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:

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