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xxTIMURxx [149]
4 years ago
9

Which two equations would be most appropriately solved by using the zero product property? Select each correct answer. 4x² = 13

0.25x2+0.8x−8=0 −(x−1)(x+9)=0 3x2−6x=0
Mathematics
2 answers:
olga55 [171]4 years ago
5 0
The zero product property tells us that if we multiply two quantities together and get zero, then one or both of these is equal to zero.

In symbols this means that if ab = 0 then a = 0, b = 0 or both.

This means that if we have a product equal to zero one of both of the factors (the things we are multiplying together) is equal to zero.

Of the equations given, the only one that has product (a few terms multiplied together) equal to zero is -(x-1)(x+9)=0 so this is the one that can most appropriately be solved by the zero product property.

We can re-write -(x-1)(x+9) = 0 as (-1)(x-1)(x+9) = 0 which means (x-1) or (x+9) or both equal zero. It should be clear that -1 cannot equal zero.

To solve we set x - 1 = 0 and obtain x = 1.
Next we set x + 9 = 0 and obtain x = -9

The solutions to the equation are x = -9 and x = 1.

We are, however, asked for two equations. Though none of the remaining equations has a product set equal to zero, we can re-write the equation 3 x^{2} -6x=0 by factoring 3x from each term on the left-hand side. Doing so gives us 3x(x-2)=0. Now we have a product equal to zero which means that one or both of the factors on the left-hand side equals zero.

That is, 3x equals 0, or x-2 equals 0 or both do. If 3x = 0 then x = 0. If x-2=0 then x=2. As such the solution to the equation is x = 0, x = 2.
netineya [11]4 years ago
5 0

Answer:

−(x−1)(x+9)=0

And..

3x2−6x=0

Step-by-step explanation:

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Which of the following could be points on the unit circle: A. (4/3, 4/5) B. (5/13, 12/13)
IRINA_888 [86]

Answer:

Option (2) and (4) are correct.

(\frac{5}{13},\frac{12}{13} ) and  (\frac{6}{7},\frac{\sqrt{13}}{7} ) are points on the unit circle.

Step-by-step explanation:

Given : Some points of circles.

We have to choose which points could be points on the unit circle.

We know, the equation of circle is

x^2+y^2=r^2  ..........(1)

where r is radius of circle.

We check each point for x and y values on the equation of circle and see which point gives radius = 1

Thus,

1) (\frac{4}{3},\frac{4}{5} )

Put  in LHS of  (1) , we have,

(\frac{4}{3})^2+(\frac{4}{5})^2

Simplify, we have,

=\frac{16}{9}+\frac{16}{25}

=\frac{544}{225}\neq 1

Thus,  (\frac{4}{3},\frac{4}{5} ) is not a point on the unit circle.

2) (\frac{5}{13},\frac{12}{13} )

Put  in LHS of  (1) , we have,

(\frac{5}{13})^2+(\frac{12}{13})^2

Simplify, we have,

=\frac{25}{169}+\frac{144}{169}

=\frac{169}{169}= 1

Thus, (\frac{5}{13},\frac{12}{13} ) is a point on the unit circle.

3) (\frac{1}{3},\frac{2}{3} )

Put  in LHS of  (1) , we have,

(\frac{1}{3})^2+(\frac{2}{3})^2

Simplify, we have,

=\frac{1}{9}+\frac{4}{9}

=\frac{5}{9}\neq 1

Thus,  (\frac{1}{3},\frac{2}{3} ) is not a point on the unit circle.

4)  (\frac{6}{7},\frac{\sqrt{13}}{7} )

Put  in LHS of  (1) , we have,

(\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2

Simplify, we have,

=\frac{36}{49}+\frac{13}{49}

=\frac{49}{49}= 1

Thus,  (\frac{6}{7},\frac{\sqrt{13}}{7} ) is a point on the unit circle.

Thus, (\frac{5}{13},\frac{12}{13} ) and  (\frac{6}{7},\frac{\sqrt{13}}{7} ) are points on the unit circle.

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