How do you find the surface area of a cube?

Mathematics

1 answer:

aalyn [17]3 years ago

7 0

A=6a^2 is how you solve the surface area of a cube

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Helppppp pleaseeeeeee

vekshin1

The answers are
AC = BD
M

4 0

2 years ago

Prove: cot(x) sec^4(x) = cot(x) +2tan(x) + tan^3(x)

masha68 [24]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad\qquad 
sec(\theta)=\cfrac{1}{cos(\theta)}\quad\qquad 1+tan^2(\theta)=sec^2(\theta)\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\
-------------------------------\\\\
cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)\\\\
-------------------------------\\\\$

$\bf \textit{so, let's do the left-hand-side}\\\\
cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)]
\\\\\\
cot(x)[1^2+2tan^2(x)+tan^4(x)]
\\\\\\
cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x)
\\\\\\
cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)}
\\\\\\
cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}$

8 0

3 years ago

In your own words describe surface and area

Oksana_A [137]

Answer:

Hi there!

Surface~ is the outside part, uppermost area

Area~ the extent or measurement of a surface

8 0

3 years ago

PLEASE HELP ME I NEED IT<br> X=<br> Y=

vampirchik [111]

X=33 and Y= 57

Step by step explanation if you see the right angle at 57 degrees it part of x so x plus 57 is 90 and 90 minus 57 is 33 so x is 33 and x and y also form a right angle so x plus y should equal 90 and 90 minus 33 is 7 so that’s Y

7 0

3 years ago

Read 2 more answers

Please answer thank u ASAP.

Neko [114]

Answer:

Step-by-step explanation:

7 0

3 years ago