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4 years ago

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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DerKrebs [107]

<h2>Hello!</h2>

The answers are:

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

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

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

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:

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

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

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

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

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

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

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

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

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So, the obtained expression is not a difference of squares

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

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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4 years ago