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vazorg [7]
4 years ago
11

What is the value of e^ ln^ 7x

Mathematics
2 answers:
goldfiish [28.3K]4 years ago
8 0

Answer:

Step-by-step explanation:

when x is greater than 0

e^{ln(7x) } = 7x

enot [183]4 years ago
3 0
It's 1/7 e 7 x. the seven and the x are exponents.
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4. Which equation describes the line that contains the points (6,2) and (12,4)?
saveliy_v [14]

Answer:

(a) y = 1/3 x is the correct equation.

Step-by-step explanation:

The given points are (6,2) and (12,4)

Now, if these points lie on any equation then they should satisfy the equation on which they lie.

CHECK FOR  equation  y = 1/3 x

if x = 6, y = 2 , then 2 = 1/3(6)  , True

if x = 12, y = 4, then 4 = 1/3(12). True

So, y = 1/3x contains both the points.

CHECK FOR  equation  y = 3x + 2

if x = 6, y = 2 , then 2 ≠ 3(6) + 2  

if x = 12, y = 4, then 4 ≠ 3(12). + 2

So, y = 3x  + 2  do not contain both the points.

CHECK FOR  equation  y = 3x  

if x = 6, y = 2 , then 2 ≠ 3(6)  

if x = 12, y = 4, then 4 ≠ 3(12).

So, y = 3x  do not contain both the point

Hence,  (a) y=1/3x is the correct equation.

8 0
4 years ago
Which of the following is an identity? A. sin2x sec2x + 1 = tan2x csc2x B. sin2x - cos2x = 1 C. (cscx + cotx)2 = 1 D. csc2x + co
Ne4ueva [31]
There are three 'Pythagorean' identities that we can look at and they are

sin²(x) + cos²(x) = 1
tan²(x) + 1 = sec²(x) 
1 + cot²(x) = csc²(x)

We can start by checking each option to see which one would give us any of the 'Pythagorean' identities as its simplest form

Option A:

sin²(x) sec²(x) + 1 = tan²(x) csc²(x)

Rewriting sec²(x) as 1/cos²(x)
Rewriting tan²(x) as sin²(x)/cos²(x)
Rewriting csc²(x) as 1/sin²(x)

We have

sin^{2}(x)[ \frac{1}{ cos^{2}(x) }]+1=[ \frac{ sin^{2}( x)}{ cos^{2} (x)}][ \frac{1}{ sin^{2}(x) } ]
[\frac{ sin^{2}(x) }{ cos^{2}(x) } ]+1= \frac{1}{ cos^{2}(x) }
tan^{2}(x)+1= sec^{2}(x)

Option B:

sin²(x) - cos²(x) = 1

This expression is already in the simplest form, cannot be simplified further

Option C:

[ csc(x) + cot(x) ]² = 1

Rewriting csc(x) as 1/sin(x)
Rewriting cot(x) as cos(x)/sin(x)

We have

[ \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}] ^{2} =1
\frac{1}{sin^2(x)}+2( \frac{1}{sin(x)})( \frac{cos(x)}{sin(x)})+ \frac{cos^2(x)}{sin^2(x)}=1csc^2(x)+2csc^2(x)cos(x)+cot^2(x)=1

Option D:

csc²(x) + cot²(x) = 1

Rewriting csc²(x) as 1/sin²(x) and cot²(x) as cos²(x)/sin²(x)

\frac{1}{sin^2(x)}+ \frac{cos^2(x)}{sin^2(x)}=1
\frac{1+cos^2(x)}{sin^2(x)} =1
1+cos^2(x)=sin^2(x)
1=sin^2(x)-cos^2(x)

from our working out we can see that option A simplified into one of 'Pythagorean' identities, hence the correct answer
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Use the cos rule

cos^-1(18^2+28^2-12^2 divided by 2*18*28

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