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2 years ago

Please help me for 40p and brainliest for 5 please

Mathematics

2 answers:

saul85 [17]2 years ago

4 0

1) The last option is your answer. It is not the same equation as the others when rearranged.

2) 1/3(12x - 24) = 16

4x - 8 = 16
4x = 16 + 8
4x = 24
x = 6, the second option.

3) Notice they are subtracting the constant term of 5 from both sides. Since it's subtraction, your answer is the second option.

4) 3/5(30x - 15) = 72

18x - 9 = 72 or 3(6x - 9) = 72
18x = 72 + 9
18x = 81
x = 81 / 18
x = 4.5

The 3rd, 4th, 5th, and 8th options are all that apply.

5) 2/3(6x + 12) = - 24

4x + 8 = - 24
4x = - 24 - 8
4x = - 32
x = - 8

The 1st, 6th, and 8th options are all that apply.

Wittaler [7]2 years ago

3 0

Answer:

1) The last option is your answer. It is not the same equation as the others when rearranged.

2) 1/3(12x - 24) = 16

4x - 8 = 16

4x = 16 + 8

4x = 24

x = 6, the second option.

3) Notice they are subtracting the constant term of 5 from both sides. Since it's subtraction, your answer is the second option.

4) 3/5(30x - 15) = 72

18x - 9 = 72 or 3(6x - 9) = 72

18x = 72 + 9

18x = 81

x = 81 / 18

x = 4.5

The 3rd, 4th, 5th, and 8th options are all that apply.

5) 2/3(6x + 12) = - 24

4x + 8 = - 24

4x = - 24 - 8

4x = - 32

x = - 8

The 1st, 6th, and 8th options are all that apply.

Step-by-step explanation:

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Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.

Tems11 [23]

Answer:

The area of the triangle is $\sqrt{3}$

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is = $\frac{1}{2}(base \times height)$

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE = $\sqrt{(0-1)^2 + (0-1)^2}$

DE = $\sqrt{(-1) ^2 + (-1)^2$

DE = $\sqrt{1+1}$

DE = $\sqrt{2}$

Base = $\sqrt{2}$

Height is DF

DF = $tan(60^{\circ}) \times DE$

DF = $\sqrt{3} \times DE$

DF = $\sqrt{3} \times\sqrt{2}$

Now, the area of the triangle is

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})$

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})$

= $\frac{1}{2}(2\sqrt{3} )$

= $\sqrt{3}$

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2 years ago

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