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Aleonysh [2.5K]
3 years ago
10

How can you use the Distributive Property to check your work when factoring? Once an expression has been factored, use the Distr

ibutive Property to expand the expression. The factored expression should be the original expression. Once an expression has been factored, use the Distributive Property to factor the expression. The factored expression should be the original expression. Once an expression has been factored, use the Distributive Property to factor the expression. The expanded expression should be the original expression. Once an expression has been factored, use the Distributive Property to expand the expression. The expanded expression should be the original expression.
Mathematics
2 answers:
yan [13]3 years ago
6 0

Answer:

 

Once an expression has been factored, use the Distributive Property to expand the expression. The expanded expression should be the original expression.

Step-by-step explanation:

i had the same question

Mariana [72]3 years ago
3 0

Answer:

Once an expression has been factored, use the Distributive Property to expand the expression. The expanded expression should be the original expression.

Step-by-step explanation:

The distributive property for any binary operation Ф under * in the given set S where a, b, c ∈ S

The binary operation a * (b Ф c) = (a * b) Ф (a * c) .

The operation * distributes over the operation Ф.

Now, when factoring, suppose we have an expression ab + ac. To factor this, we have a(b + c).

To check that our answer is correct, since the expression has been factored, we then distribute the multiplication over the addition to check if the factorization is correct.

So, a(b + c) = ab + ac which is our initial expression.

So, to use the distributive property to check work when factoring, once an expression has been factored, use the Distributive Property to expand the expression. The expanded expression should be the original expression.

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The correct option is;

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Lina a passes through (-1, 0) and (1, 2)

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Line d passes through y = 1

An equation that has no solution

By examination, from the plot of the coordinates using an online tool, the system that would have no solution is the system of line a and line b because they are parallel lines and cannot meet or have a point where there is a common solution, therefore, there are no solutions when the equations of the two lines are combined in one equation

Slope of line a = -2/-2 = 1

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The slopes of lines a and b are equal, therefore, line a and line b are parallel.

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A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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