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alexira [117]
3 years ago
10

...................................................

Mathematics
1 answer:
Verizon [17]3 years ago
6 0

Answer:

Answer

.....................................................

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What is three-fourths of 1,968
lara31 [8.8K]
All you have to so is divide 1968 by 4.

1968 ÷ 4 = 496.5 = 1/4


Now multiply 496.5 by 3 =  1, 489.5 = 3/4


I hope this helps! :_
7 0
3 years ago
Read 2 more answers
Solve for p<br> 3+|1+p|=12
Usimov [2.4K]
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 &amp; Simplify.
kipiarov [429]
<span>2a(7a²-3a+9)
=14a^3 - 6a^2 +18a

answer is </span><span>D. 14a³-6a²+18a</span>
3 0
3 years ago
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