Which is the equivalent of 12.42 written in DMS form?

Mathematics

2 answers:

scZoUnD [109]3 years ago

7 0

Answer:

DMS form = 12°25'12''.

Step-by-step explanation:

Given : 12.42.

To find : written in DMS form.

Solution : We have given that 12.42.

Multiply 42 by 60 to convert it in to minute

We can write it as 12 + .42(60).

12°25.2'

Rewrite 12°25+.2'

2' = 2 ( 60) to convert minute in to second.

12°25+.2(60)

12°25'12''

Therefore, DMS form = 12°25'12''.

nika2105 [10]3 years ago

3 0

12.42

12°+.42(60)

12°25.2'

12°25'+.2(60)

12°25'12"

You might be interested in

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72