5-6(a+2)=7+a What is a

2 answers:

iogann1982 [59]2 years ago

5 0

Remmeber you can do anything to an equaiton as long as you do it to btoh sides
5-6(a+2)=7+a
distribute
5-6a-12=7+a
add like terms
-6a-7=7+a
add 6a both sides
-7=7+7a
minus 7 both sides
-14=7a
divide by 7
-2=a
a=-2

morpeh [17]2 years ago

4 0

The first step here would be to distribute the parenthesis to the -6.

5 - 6a - 12 = 7 + a

Next, combine all like terms. This means that the constants will combine, thus making the next step easier.

- 6a - 7 = 7 + a

Now, subtract the negative seven from the other side of the parenthesis. When I say add the negative, I mean that the sign (-) will flip, turning into a positive.

- 6a = 14 + a

Now do the same with the a, except this time the a will be subtracted from the other side due to it's positive sign being flipped into a negative.

- 7a = 14

Now you divide both sides by - 7. Almost done!!!

- 7a / - 7 = 14 / -7

Finally the equation simplifies to:

a= -2

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First, I add 1. Then, I multiply by 2. Next, I divide by 3. As a result, I get 4. What is my starting number?

Ira Lisetskai [31]

Answer:5

Step-by-step explanation:

reverse the steps

2(x+1)/3=4

2(x+1)/3×3=4×3

2(x+1)/2=12/2

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6 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

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