All categories

3 years ago

What is 6+3(-8x-2)=12

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

3 0

Answer:

6 + 3 (-8 x -2)

Step-by-step explanation:

6 + 3 (-8 x -2)

= 6+3 ( 16 )

= 9 ( 16 )

= 144

You might be interested in

HELP HELP PLEASE IM BEGGING YOUUU

Alexandra [31]

Answer:

Area=615.75

Step-by-step explanation:

A=πr2

Seriously I have helped you twice in 5 minutes.

8 0

3 years ago

the width of a rectangle is 5 inches and the height is 10 inches what is the perimeter of the shaded rectangle

andrew-mc [135]

So, you should add the heighth twice and add the width twice too, then add those two together to get your answer.

7 0

3 years ago

Read 2 more answers

PLZ HELPPPPPPPPPP<br>IT UA DUE IN 30 MINUTES

LenKa [72]

Answer:

yes

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

For what value of x is line a parallel to line b?<br><br> (Yo, please help me so I can go to sleep.)

iogann1982 [59]

X=20

If they're parallel the other number in the same spot as 5x+15 equals that, so 5x+15=115 because if they are parallel they're the same if same spot and same line going through them.

115-15 = 100
5x=100
Divide by 5 both sides
X=20

If x=20 then these lines would be parallel

6 0

3 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago