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serious [3.7K]
3 years ago
12

Someone please help me It’s getting late and it’s due tomorrow

Mathematics
1 answer:
inessss [21]3 years ago
4 0

Answer:

After the distribution of the 5 and the addition of x to both sides, the x variable is cancelled out leaving 1 = - 20.

This statement is false. Hence, no solution

Step-by-step explanation:

1 - x = 5(x - 4) - 6x (Distribute 5)

1 - x = 5x - 20 - 6x (Combine like terms)

1 - x = -x - 20  add x to both sides

1 = - 20 False

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Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if
Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula, we get

JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}

JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}

JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}

JK=\sqrt{4+9}

JK=\sqrt{13}

Similarly,

KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6

LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}

JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6

Now, perimeter P of ▱JKLM is

P=JK+KL+LM+JM

P=\sqrt{13}+6+\sqrt{13}+6

P=2\sqrt{13}+12

P=2(3.61)+12

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P=19.22

P\approx 19.2

Therefore, the perimeter P of ▱JKLM is 19.2 units.

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Answer:

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Step-by-step explanation:

It is given that there are two triangles \triangleABC and

\triangleABC ~

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∠B≅∠E

Please have a look at the attached figure for \triangleABC and

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