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Ostrovityanka [42]
3 years ago
6

Find the critical numbers of f(x)=2secx+tanx in 0

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
3 0
Using the range of [ 0 , 2π ]


\bf f(x)=2sec(x)+tan(x)\implies \cfrac{df}{dx}=2sec(x)tan(x)+sec^2(x)
\\\\\\
0=2sec(x)tan(x)+sec^2(x)
\\\\\\
-sec^2(x)=2sec(x)tan(x)\implies \cfrac{-sec^2(x)}{2sec(x)}=tan(x)
\\\\\\
\cfrac{-sec(x)}{2}=tan(x)\implies \cfrac{-\frac{1}{cos(x)}}{\frac{2}{1}}=\cfrac{sin(x)}{cos(x)}
\\\\\\
-\cfrac{1}{2cos(x)}=\cfrac{sin(x)}{cos(x)}\implies -\cfrac{1}{2}=sin(x)\implies \measuredangle x =
\begin{cases}
\frac{7\pi }{6}\\\\
\frac{11\pi }{6}
\end{cases}
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Answer:

E is not a subspace of \mathbb{R}^2

Step-by-step explanation:

E is not a subspace of  \mathbb{R}^2

In order to see this, we must find two points (a,b), (c,d) in  E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.  

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

5 0
3 years ago
If 9 is added to twice a number and this sum is multiplied by 3​, the result is the same as if the number is multiplied by 4 and
Lina20 [59]
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4 0
3 years ago
How do you solve -5+-3?
Paha777 [63]
- 5 + -3 = -8

hope this helps


when a negative number adds a negative number, it will be negative

when a negative number adds a positive number, whichever number (either positive or negative) take the sign of the bigger one

For example:

-4 + 3 = -1

-5 + 10 = 5



hope this helps
8 0
3 years ago
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Step-by-step explanation:

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8 0
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