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laila [671]
2 years ago
8

Write the equation of a line that passes through (3,-2) and (-6, 11) in standard form.

Mathematics
1 answer:
gtnhenbr [62]2 years ago
8 0
Y = mx+c

m = 11 - (-2) / -6-(-3) = -13/9

Substitute into equation

11 = -13/9(-6)+C
C = 7/3

y = -13/9x + -13/3 multiply with 9 n you will get

Answer = y = -13x - 39
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7/9 - 2/3 and 2/3 - 1/6 <br>​
arsen [322]

Answer:

The answer is 1/9 and 1/2

7 0
3 years ago
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How do I do this?<br>*Look at the directions in the photo*​
lora16 [44]

Answer:

Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2

Step-by-step explanation:

We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\

As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2

For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2

For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2

Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2

7 0
2 years ago
The sum of the two angles of a triangle is 80 degree and their difference is 20 degree. find all the angles of the triangle.
Norma-Jean [14]

Sum of 2 angles = 80°

Difference = 20°

If we take away 20° from the sum, both angles are equal

⇒80 - 20 = 60°


Divide by 2 to find the smaller angle:

60 ÷ 2 = 30


Add 20° to find the bigger angle:

20 + 30 = 50°


<u>One of the angle is 30° and the other angle is 50°</u>


Given that the sum of the two angles is 80°

⇒ Third angle = 180 - 80 = 100°


Answer: The 3 angles are 30° , 50° and 100°

3 0
3 years ago
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Solve equation 3x+3y+6z-15w=9​
Allushta [10]

Answer:

The equation is already solved.

8 0
3 years ago
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Please help me on this math please
makkiz [27]
1=60
2=120
3=120
4=180
5=240
6=300
and you have arrived bu now. :)
Good luck with your lessons, hope I helped. :D
8 0
3 years ago
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