Subtract 3 from both sides

simplify 12 - 3 to 9

break down the problem into these two equations

1 + p = 9 and -(1 + p) = 9

solve the first equation 1 + p = 9 and that would be 8 since 1 + 8 = 9 is true.

solve the second equation -(1 + p) = 9 and just simplify brackets and add 1 to both sides then add 9 + 1 and lastly multiply both sides by -1 and p = -10.

Gather both solutions

Answers: p = -10, 8

5 0

3 years ago

Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?

vichka [17]

Answer:

The expression $-1+14i$ represents the number $2i^4-5i^3+3i^2+\sqrt{-81}$ rewritten in a+bi form.

Step-by-step explanation:

The value of $i$ is $i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4}$ in term of $i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}$

Substituting the value,

$i^{4}=\left(-1\right)\times \left(-1\right)$

Product of two negative numbers is always positive.

$\therefore i^{4}=1$

Now $i^{3}$ in term of $i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i$

Substituting the value,

$i^{3}=\left(-1\right)\times i$

Product of one negative and one positive numbers is always negative.

$\therefore i^{3}=-i$

Now $\sqrt{-81}$ can be written as follows,

$\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}$

Applying radical multiplication rule,

$\sqrt{ab}={\sqrt{a}}\sqrt{b}$

$\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}$

Now, $\sqrt{\left(81\right)=9$ and $\sqrt{-1}}=i$

$\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i$

Now substituting the above values in given expression,

$2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i$

Simplifying,

$2+5i-3+9i$

Collecting similar terms,

$2-3+5i+9i$

Combining similar terms,

$-1+14i$

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

5 0

3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

4 0

3 years ago

Multiply & Simplify.

kipiarov [429]

2a(7a²-3a+9)
=14a^3 - 6a^2 +18a

answer is D. 14a³-6a²+18a

3 0

3 years ago