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Marta_Voda [28]
3 years ago
7

6. Select all the inequalities that have the same solutions as -4x < 20.

Mathematics
1 answer:
attashe74 [19]3 years ago
7 0

Answer:

F and B (as far as I can see).

Step-by-step explanation:

First, let's simplify the equation. You can easily do this by dividing by -4.

                                                  -4x < 20

                                                   /-4    /-4

                                                      x > -5

REMEMBER- Whenever dividing or multiplying a negative number with inequalities, the sign flips!!

On A, we can see that it has a negative symbol, but if we were to divide that, the sign would flip, making -5 greater than x. That's not what we want.

On B, the 20 is negative instead of the 4. Does this change anything?

                                                   4x > -20

                                                   /4    /4

                                                      x > -5

Nope! It still simplifies to what we had at the beginning so it works!

C and D are slightly confusing me because, well, I don't really see how it's relevant to anything so... It doesn't even have a inequality symbol so I don't know lol.

E is wrong because we need the 5 to be positive.

F is obviously right compared to our first answer.

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1) The variables are "k"

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Answer:

Perimeter \sqrt{26}+\sqrt{5}+\sqrt{10}+\sqrt{17} units.  Area 12 square units.

Step-by-step explanation:

Perimeter: total distance around the figure.

Distance Formula:  the distance between points \left(x_1,y_1\right) \text{ and } \left(x_2,y_2\right) is

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

AB=\sqrt{(6-1)^2+(2-1)^2}=\sqrt{25+1}=\sqrt{26}

BC=\sqrt{(5-6)^2+(4-2)^2}=\sqrt{1+4}=\sqrt{5}CD=\sqrt{(2-5)^2+(5-4)^2}=\sqrt{9+1}=\sqrt{10}DA=\sqrt{(1-2)^2+(1-5)^2}=\sqrt{1+16}=\sqrt{17}

The perimeter is the sum of all those segment lengths.

One way to find the area of the figure is to surround it with a rectangle, insert some lines so that the areas you do not want can be found and subtracted from the rectangle's area.  (See attached image.)

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A. (1/2)(1)(4) = 2 square units

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D. (1/2)(1)(2) = 1 square unit

E. (1/2)(5)(1) = 2.5 square units

Square C.  (1)(1) = 1 square unit

Total of all the area you don't want to include:

2 + 1.5 + 1 + 2.5 + 1 = 8 square units

Subtract 8 from the surrounding rectangle's area of 20, and you get the area of the figure is 20 - 8 = 12 square units.

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