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lubasha [3.4K]
3 years ago
14

Plese help i find this confusing no links :(

Mathematics
2 answers:
aleksklad [387]3 years ago
8 0

Answer:

e 5,3

Step-by-step explanation:

wolverine [178]3 years ago
8 0
The answer too this problem is E
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Verify the identity cot(x-pi/2)=-tan x
vampirchik [111]
To verify the identity we need the following identities:

i) \displaystyle{ \cot(x)= \frac{\cos x}{\sin x}

ii) \displaystyle{ \sin (x-y)=\ sinx\cdot\ cosy -\ siny\cdot\ cosx

iii) \displaystyle{ \cos (x-y)=\ cosx\cdot \ cosy +\ sinx\cdot\ siny.

Also, we have know that \displaystyle{ \sin \frac{ \pi }{2}=1 and \displaystyle{ \cos \frac{ \pi }{2}=0.


Thus, \displaystyle{ \cot(x-\frac{\pi}{2})= \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}

By (ii) and (iii) we have:

\displaystyle{ \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}= \frac{\ cosx\cdot \ cos\frac{\pi}{2} +\ sinx\cdot\ sin\frac{\pi}{2}}{\ sinx\cdot\ cos\frac{\pi}{2} -\ sin\frac{\pi}{2}\cdot\ cosx} = \frac{\ sinx}{-\cos x}

by simplifying \displaystyle{ \sin \frac{ \pi }{2}=1 and \displaystyle{ \cos \frac{ \pi }{2}=0.

Now, \displaystyle{ \frac{\ sinx}{-\cos x} is clearly -tanx.
4 0
3 years ago
How can i complete whole maths syllabus in 7 days of exam ?​
alex41 [277]
Study? I’m not sure what this is asking
5 0
3 years ago
Read 2 more answers
10 POINTS!!! FULL ANSWER WITH FULL STEP BY STEP SOLUTION PLEASE. DO BOTH PARTS OF 1 AND ALL OF 2.
diamong [38]
One A
y = e^x
dy/dx = e^x The f(x) = the differentiated function. Any value that e^x can have, the derivative has the same value. x is contained in all the reals.
One B
y = x*e^x
y' = e^x + xe^x Using the multiplication rule.
You want the slope and the value of the of y to be the same. The slope is y' of the tangent line
xe^x = e^x + xe^x
e^x = 0
This happens only when x is very "small" like x = - 4444444

y = e^x * ln(x) Using the multiplication rule again, we need the slope of the line with is y'
y1 = e^x
y1' = e^x
y2 = ln(x)
y2' = 1/x
y' = e^x*ln(x) + e^x/x So at x = 1 the slope of the line = 
y' = e^1*ln(1) + e^1/1
y' = e*0+e  = e
y = mx + b
y = ex + b
to find b we use y= e^x ln(x)

e^x ln(x) = e*x + b
e^1 ln(1) = e*1 + b
ln(1) = 0
0 = e + b
b = - e

line equation and answer.
y = e*x - e
8 0
3 years ago
The area of a rectangular wall of a barn is 20 square feet. Its length is 8 feet longer than the width. Find the length and widt
Margarita [4]

Answer:

It there a picture

Step-by-step explanation:

4 0
3 years ago
For an integer x,x squared minus x will always produce an even value
yuradex [85]

Answer:

Step-by-step explanation:

Not exactly sure what your question is - I am assuming that it is something like:

Show/prove that for any integer x, x^2 - x is even.

Suppose that x is an even integer.  The product of an even integer and any other integer is always even (x = 2n, so x * y = 2 n * y which is even.  Therefore x^2 is even.  An even minus an even is even.  (The definition of an even number is that it is divisible by 2 or has a factor of 2.  So the difference of even numbers could be written as 2*( the difference of the two numbers divided by 2); therefore the difference is even)

Suppose that x is an odd integer.  The product of 2 odd numbers is odd - each odd number can be written as the sum of an even number and 1; multiplying the even parts with each other and 1 will produce even; multiplying the 1's will produce 1, so the product can be written as the sum of an even number and 1 - which is an odd number.  The difference between two odd numbers is even - the difference between the even parts is even (argument above), the difference between 1 and 1 is zero, so the result of the difference is even.

x^2 is therefore even if x is even and odd if x is odd; The difference x^2 - x is even by the arguments above.

5 0
3 years ago
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