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meriva
2 years ago
7

Tan2x=___

Mathematics
1 answer:
SSSSS [86.1K]2 years ago
7 0

Step-by-step explanation:

tan2x = sin2x/cos2x

we know that ,

  • sin2x = 2sinxcosx
  • sin2x = 2sinxcosx cos2x = cos²x - sin²x

Since, tan2x = 2sinx*cosx/cos²x- sin²x

Dividing by cos²x in numerator and denominator.

We get ,

  • tan2x = 2sinx/cosx / 1- sin²x/cos²x

  • Since, we get tan2x= 2tanx/1-tan²x

  • Option (c) is correct option.

  • Hope it will be helpful
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If 2x + 6y = 16, which of the following expressions represents x in terms of y?
Andrej [43]

Answer:

x=-3y+8

Step-by-step explanation:

2x+6y=16

2x=16-6y

2x=-6y+16

x=-6/2y+16/2

x=-3y+8

7 0
3 years ago
Y=(3/5)x -3 and 5y =3x-10 these lines are and the second question?
azamat
Look at the first line:  y = (3/5)x - 3.  What happens if you multiply each term by 5, to eliminate the fraction?

5y = 3x - 3

Compare this to the second equation, 

5y - 3x = -10, or 5y = 3x - 10.

The coefficients of x and y (as 3 and 5 here) determine the slope of a straight line.  Since 5y = 3x is present in both equations, the two lines MUST be parallel.



y = 4
4y = 6   =>   y = 6/4

y+4 and y =3/2 are both horizontal lines.  Since they are horiz., they are parallel.


4 0
3 years ago
The scatter plot shows how the cooking time for a beef roast is related to its weight. Which statement is NOT supported by the t
bija089 [108]

Answer:

the statement that is not supported by the trend in this scatter plot is that each additional pound adds about 1 hour of cooking time. (last option)

3 0
3 years ago
Read 2 more answers
Please helpppp<br><br> Extra points and brainleist
vladimir1956 [14]

Answer:

B

Step-by-step explanation:

x² - 4x - 7 = 0

(x - 2)² = 11

5 0
3 years ago
If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
3 years ago
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