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3 years ago

2x divided by x+5 less than or equal to 0

Mathematics

1 answer:

coldgirl [10]3 years ago

4 0

It is more than 0, because you cancel the x, and you get 2+5, which is 7

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Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

3 years ago

Find the x- and y-intercepts of 5x - 3y = 12 .

irga5000 [103]

X = (12/5,0)
y = (0,-4)

substitute 0 in for y & solve for x , to find the y-intercept substitute 0 in for x and then solve . ( I hope that makes sense)

6 0

3 years ago

Determine whether each equation is a linear equation. Write yes or no. If yes, write the equation in standard form. y = 4x + x

Step2247 [10]

Answer:

The answer is y = 5x

Step-by-step explanation:

Yes, this equation is a linear equation.

because it has degree 1

Standard Form

y = 4x + x

y = 5x

Thus, The answer is y = 5x

-TheUnknownScientist 72

5 0

2 years ago

2:x and 12:18 a-3, b-4, c-6

nirvana33 [79]

The Corrected Problem is :

2:x and 12:18 identify the value of x that makes each pair of ratios equivalent .

Solution:

If a pair of Ratios are equivalent then we can write

$\frac{2}{x}=\frac{12}{18}\\ \\ \text{Simplify we get}\\ \\ 12x=36\\ \\ \text{Divide both the sides by 12 we get}\ \\ \\ \frac{12x}{12}=\frac{36}{12}\\ \\ x=3\\ \\ \text{Hence the required value of x is 3}\\$

3 0

3 years ago

Read 2 more answers

PLZ HELP ASAP DIVIDE LINE SEGMENTS

rodikova [14]

Segment in the direction from A to C
Initial Point: A=(9,5)=(xa,ya)→xi=xa=9, yi=ya=5
Final point: C=(-7,1)=(xc,yc)→xf=xc=-7, yf=yc=1
B=(xb,yb)=?
Proportion: r=AB/BC=3:1=3/1→r=3

xb=(xi+r*xf)/(1+r)
Replacing xi=xa=9, xf=xc=-7 and r=3
xb=[9+3*(-7)]/(1+3)
xb=(9-21)/4
xb=(-12)/4
xb=-3

yb=(yi+r*yf)/(1+r)
Replacing yi=ya=5, yf=yc=1 and r=3
yb=[5+3*(1)]/(1+3)
yb=(5+3)/4
yb=8/4
yb=2

B=(xb,yb)→B=(-3,2)

Answer: B=(-3,2)

7 0

3 years ago