WILL MARK BRAINLIEST IF ANSWER IS CORRECT

Mathematics

1 answer:

crimeas [40]3 years ago

6 0

The last set (choice D) is a subset of given set B.

The other 3 choices are wrong.

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13 add 2x equals 3 what does x equal??

Anastasy [175]

13 + 2x = 3
subtract 13 from both sides of the equation
2x = -10
divide by 2 on both sides
x = -5

5 0

2 years ago

Read 2 more answers

Solve each quadratic equation. Show your work 15. x2 + 3x = 10

OLEGan [10]

X² + 3x = 10

Convert to standard form.
x² + 3x - 10 = 0

factor x² = x * x
factor -10 = 5 * -2

(x + 5) (x - 2) = 0
x(x -2) + 5(x-2) = 0
x² - 2x + 5x - 10 = 0
x² + 3x - 10 = 0

x + 5 = 0   x² + 3x = 10
x = -5 -5² + 3(-5) = 10
25 - 15 = 10
10 = 10

x - 2 = 0 x² + 3x = 10
x = 2 2² + 3(2) = 10
   4 + 6 = 10
   10 = 10

4 0

3 years ago

Read 2 more answers

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:

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

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

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

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

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