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GarryVolchara [31]
3 years ago
9

Factor 28+56t+28w28+56t+28w28, plus, 56, t, plus, 28, w to identify the equivalent expressions. Choose 2 answers:

Mathematics
1 answer:
Effectus [21]3 years ago
6 0

Answer: The factorization of  28+56t+28w =28(1+2t+w)

The factors of 28+56t+28w are 28 and 1+2t+w.

Step-by-step explanation:

The given expression : 28+56t+28w

To factorize it, we need to find the common factor.

As 56 can be written as 2 x 28.

So, the above expression would become

28+2\times28t+28w

Now, taking 28 as common from all the terms, we will get

28(1+2t+w)

Thus, the factorization of  28+56t+28w =28(1+2t+w)

And the factors of  28+56t+28w are 28 and 1+2t+w.

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Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE
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Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to 20\ units

Part 3) The area is equal to 25\ units^{2}

Step-by-step explanation:

we have

A(5,0), B(2,4), C(-2,1),D(1,-3)

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see the attached figure

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance AB

A(5,0),B(2,4)

substitute in the formula

d=\sqrt{(4-0)^{2}+(2-5)^{2}}

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Find the distance BC

B(2,4), C(-2,1)

substitute in the formula

d=\sqrt{(1-4)^{2}+(-2-2)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}

BC=5\ units

Find the distance CD

C(-2,1),D(1,-3)

substitute in the formula

d=\sqrt{(-3-1)^{2}+(1+2)^{2}}

d=\sqrt{(-4)^{2}+(3)^{2}}

d=\sqrt{25}

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Find the distance AD

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substitute in the formula

d=\sqrt{(-3-0)^{2}+(1-5)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}        

AD=5\ units

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

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substitute in the formula

d=\sqrt{(-3-4)^{2}+(1-2)^{2}}

d=\sqrt{(-7)^{2}+(-1)^{2}}

BD=\sqrt{50}\ units        

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

BD^{2}=AD^{2}+AB^{2}

substitute

(\sqrt{50})^{2}=5^{2}+5^{2}

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The triangle BDA is a right triangle

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The area of a square is equal to

A=b^{2}

we have

b=5\ units

A=5^{2}=25\ units^{2}

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

P=4b

we have

b=5\ units

P=4(5)=20\ units

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Step-by-step explanation:

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