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4 years ago

Evaluate the following: −3 − (−2). (5 points) Question 1 options: 1) −5 2) −1 3) 1 4) −6

Mathematics

1 answer:

Ganezh [65]4 years ago

7 0

Answer:

2) -1

Step-by-step explanation:

it is easy

we have to first open the brackets, as it is ' - ' we have to change the sign to opposite of that of the number in the bracket and then solve it.

so the number 2 is having negative sign, we make it positive. this rule is applied only when there is a negative sign and a bracket is there.

-3-(-2)

-3+2

=-1

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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

$P=JK+KL+LM+JM$

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

$P=2\sqrt{13}+12$

$P=2(3.61)+12$

$P=7.22+12$

$P=19.22$

$P\approx 19.2$

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

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