All categories

2 years ago

If one side of a square measures 6 units, what is its area?

Mathematics

2 answers:

aniked [119]2 years ago

8 0

Answer: Area=36

Step-by-step explanation:

Harrizon [31]2 years ago

4 0

Answer:

36 units²

Step-by-step explanation:

By the definition of a square, all sides of the square has the same measurement.

Area of a square = side x side.

If a side = 6 units; Plug in 6 for side:

Area (A) = side x side

A = 6 x 6

A = 36

Your area is 36 units².

You might be interested in

Pani-rosa [81]

Answer:

i thinks it's b

Step-by-step explanation:

I do not no tho

3 0

3 years ago

Which identify could be used to rewrite the expression x6-27

hammer [34]

The differnce of 2 perfect cubes

remember
a³-b³=(a-b)(a²+ab+b²)

so
$(x^2)^3-(3)^3=(x^2-3)(x^2+3x+9)$

3 0

3 years ago

Given the following formula, solve for v.<br> s = 1/2a^2v + c

weeeeeb [17]

Since there is no initial condition, the exact value of v cannot be determined, but you can set the equation up for a general evaluation of the situation.

s = 1/2a^2*v+c

First you need to subtract c from both sides to get

s-c = 1/2a^2*v

then you can just divide both sides by 1/2a^2 to get v

(2(s-c))/a^2=v

when dividing a fraction, such as 1/2, make sure to keep in mind that you're really multiplying the flipped version, so that dividing by 1/2 means multiplying by 2.

7 0

3 years ago

Read 2 more answers

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

Please help me, I have minimum time

natita [175]

X= 5x Y=-6 I just added them together

8 0

2 years ago