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irakobra [83]
3 years ago
7

For which value of a will the equation 3(x-2)=a+3x have an infinite number of solution. A) 2 B) -6 C)6 D) -2

Mathematics
2 answers:
andrew11 [14]3 years ago
4 0
I think is C or B
Because
3(x-2) = a+ 3x
3x 6 = a + 3x
3x+3x= 6
So
6=a + 6x
Neko [114]3 years ago
3 0
I think the answer is
-6
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What is the image of(8,12) after a dilation by a scale factor of 1/4 centered at the origin?
max2010maxim [7]
<h3>Answer:  (2, 3)</h3>

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

Explanation:

1/4 = 0.25 is the scale factor

Multiply this with each coordinate of the given point

0.25*8 = 2 is the new x coordinate

0.25*12 = 3 is the new y coordinate

So (8,12) moves to (2,3) after applying the dilation

The scale factor k makes 0 < k < 1 true, so the point is closer to the origin after applying the dilation.

--------

Side note: this trick of multiplying the scale factor by each coordinate only works if the dilation is centered at the origin. For any other center, you'll need to apply a translation first, dilate, then translate back again.

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3 years 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\\\\&#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

6 0
2 years ago
Can somebody help me with this problem?
Dmitriy789 [7]
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Fittoniya [83]

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2 years ago
Find sin 2a and cot 2a:
geniusboy [140]

Answer:

sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cot(2\alpha ) = \frac{16511}{18720}

Step-by-step explanation:

\text{if } cos(\alpha)=\frac{12}{13}\\\text{That must mean we have a triangle with base 12, and hypotenuse 13.}\\\text{Using Pythagoras, we can determine the base of the triangle must be 5.}\\a^2+b^2=c^2 \text{, where c is the hypotenuse and a, b are the two other sides.}\\c^2-b^2=a^2\\\sqrt{c^2-b^2}=a\\\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\\\text{Therefore, }sin(\alpha) = \frac{5}{12}\\sin(2\alpha)=2sin(\alpha )cos(\alpha)\\\text{(From double angle formulae identities)}\\

sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cos(2\alpha )=cos^2(\alpha)-sin^2(\alpha)\\cos(2\alpha )=(\frac{12}{13})^2-(\frac{5}{12})^2=\frac{16511}{24336}\\cot(2\alpha)=\frac{cos(2\alpha)}{sin(2\alpha)}=\frac{\frac{16511}{24336}}{\frac{10}{13}}=\frac{16511}{18720}

8 0
3 years ago
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