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poizon [28]
3 years ago
14

Find u*v if u=14j and v=i+4j

Mathematics
1 answer:
8090 [49]3 years ago
6 0
\vec{u} and \vec{v} are vectors and:

 \vec{u}=14j=0i+14j=[0,14]\\\\
\vec{v}=i+4j=1i+4j=[1,4]

\vec{u}\cdot\vec{v} is the dot product or scalar product, so:

\vec{u}\cdot\vec{v}=[0,14]\cdot[1,4]=0\cdot1+14\cdot4=56

Correct answer is A.
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What are all the values of x, for which the square of x is equal to the value of x increased by 2?
Sedbober [7]
=±22
x
=
±
2
x
2


Using the fact that 2=ln2
2
=
e
ln
⁡
2
:

=±ln22
x
=
±
e
x
ln
⁡
2
2


−ln22=±1
x
e
−
x
ln
⁡
2
2
=
±
1


−ln22−ln22=∓ln22
−
x
ln
⁡
2
2
e
−
x
ln
⁡
2
2
=
∓
ln
⁡
2
2


Here we can apply a function known as the Lambert W function. If =
x
e
x
=
a
, then =()
x
=
W
(
a
)
.

−ln22=(∓ln22)
−
x
ln
⁡
2
2
=
W
(
∓
ln
⁡
2
2
)


=−2(∓ln22)ln2
x
=
−
2
W
(
∓
ln
⁡
2
2
)
ln
⁡
2


For negative values of
x
, ()
W
(
x
)
has 2 real solutions for −−1<<0
−
e
−
1
<
x
<
0
.

−ln22
−
ln
⁡
2
2
satisfies that condition, so we have 3 real solutions overall. One real solution for the positive input, and 2 real solutions for the negative input.

I used python to calculate the values. The dps property is the level of decimal precision, because the mpmath library does arbitrary precision math. For the 3rd output line, the -1 parameter gives us the second real solution for small negative inputs. If you are interested in complex solutions, you can change that second parameter to other integer values. 0 is the default number for that parameter.
8 0
3 years ago
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