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Naya [18.7K]
4 years ago
11

Select the correct answer from the drop-down menu. Find the solution set. The solution set for 5v2 – 125 = 0 is .

Mathematics
2 answers:
kozerog [31]4 years ago
5 0

your answer should be (-5,5) sorry if im wrong tho and if im wrong plz correct me

Digiron [165]4 years ago
5 0

Answer:

The solution set is v=5,-5.                                  

Step-by-step explanation:

Given : Expression 5v^2-125=0

To find : The solution set?

Solution :

Step 1 - Write the expression

5v^2-125=0

Step 2 - Add 125 both side,

5v^2-125+125=125

5v^2=125

Step 3 - Divide by 5 both side,

\frac{5v^2}{5}=\frac{125}{5}

v^2=25

Step 4 - Taking root both side,

\sqrt{v^2}=\sqrt{25}

v=\pm5

Therefore, The solution set is v=5,-5.

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<h2>Hello!</h2>

The answers are:

C) (3+xz)(-3+xz)

D) (y^2-xy)(y^2+xy)

F) (64y^2+x^2)(-x^2+64y^2)

<h2>Why?</h2>

To know which of the products results in a difference of square, we need to remember the difference of squares from:

The difference of squares form is:

(a+b)(a-b)=a^{2}-b^{2}

So, discarding each of the given options in order to find which products result in a difference of squares, we have:

<h2>A)</h2>

(x-y)(y-x)=xy-x^{2}-y^{2} +yx=-x^{2} -y^{2}

So, the obtained expression is not a difference of squares.

<h2>B)</h2>

(6-y)(6-y)=36-6y-6y+y^{2}=y^{2}-12y+36

So, the obtained expression is not a difference of squares.

<h2>C)</h2>

(3+xz)(-3+xz) =(xz+3)(xz-3)=(xz)^{2}-3xz+3xz-(3)^{2}\\\\(xz)^{2}-3xz+3xz-(3)^{2}=(xz)^{2}-(3)^{2}

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

<h2>D)</h2>

(y^2-xy)(y^2+xy)=(y^{2})^{2}+y^{2}*xy-y^{2}*xy-(xy)^{2} \\\\(y^{2})^{2}+y^{2}*xy-y^{2}*xy-(xy)^{2}=(y^{2})^{2}-(xy)^{2}

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

<h2>E)</h2>

(25x-7y)(-7y+25x)=-175xy+(25x)^{2}+49y^{2}-175xy\\\\-175xy+(25x)^{2}+49y^{2}-175xy=(25x)^{2}+49y^{2}-350xy

So, the obtained expression is not a difference of squares

<h2>F)</h2>

(64y^2+x^2)(-x^2+64y^2)=(64y^2+x^2)(64y^2-x^2)\\\\(64y^2+x^2)(64y^2-x^2)=(64y^{2})^{2} -(x^{2}*64y^{2})+(x^{2}*64y^{2})-(x^{2})^{2}\\ \\(64y^{2})^{2} -(x^{2}*64y^{2})+(x^{2}*64y^{2})-(x^{2})^{2}=(64y^{2})^{2}-(x^{2})^{2}

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

Hence, the products that result in a difference of squares are:

C) (3+xz)(-3+xz)

D) (y^2-xy)(y^2+xy)

F) (64y^2+x^2)(-x^2+64y^2)

Have a nice day!

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