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3 years ago

Which of the following is a solution to 2sin^(2)x+sinx-1=0

Mathematics

1 answer:

lukranit [14]3 years ago

6 0

Hello,

$2*sin^2(x)+sin(x)-1=0\\\Delta=1+4*2*1=9=3^2\\sin^2(x)=\frac{-1-3}{4} =-1\ (impossible)\\or\\sin^2(x)=\frac{-1+3}{4} =\dfrac{1}{2}\\sin(x)=\dfrac{\sqrt{2} }{2} \Longrightarrow (x=45^o\ or\ x=135^o)\\or\\sin(x)=-\dfrac{\sqrt{2} }{2} \Longrightarrow (x=315^o\ or\ x=225^o)\\$

Answer A : 45°

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5x+18/8 = x/4 <br> Solve for x

ollegr [7]

Answer: x= -9/19

Step-by-step explanation:

5x+18/8=x/4

5x+ 9/4=x/4

5x+9/4-9/4=x/4-9/4

5x= x-9/4

5x(4)=x-9/4(4)

20x=x-9

20x-x=x-9-x

19x=-9

19x/19=-9/19

x=-9/19

8 0

3 years ago

What is the equation, in slope-intercept form, of the line parallel to y = 5x+2 that passes through the point (-2, 1)?

o-na [289]

Y=5x+11 using the point-slope formula and simplifying

4 0

2 years ago

Show tan(???? − ????) = tan(????)−tan(????) / 1+tan(????) tan(????)<br> .

anyanavicka [17]

Answer:

See the proof below

Step-by-step explanation:

For this case we need to proof the following identity:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

We need to begin with the definition of tangent:

$tan (x) =\frac{sin(x)}{cos(x)}$

So we can replace into our formula and we got:

$tan(x-y) = \frac{sin(x-y)}{cos(x-y)}$ (1)

We have the following identities useful for this case:

$sin(a-b) = sin(a) cos(b) - sin(b) cos(a)$

$cos(a-b) = cos(a) cos(b) + sin (a) sin(b)$

If we apply the identities into our equation (1) we got:

$tan(x-y) = \frac{sin(x) cos(y) - sin(y) cos(x)}{sin(x) sin(y) + cos(x) cos(y)}$ (2)

Now we can divide the numerator and denominato from expression (2) by $\frac{1}{cos(x) cos(y)}$ and we got this:

$tan(x-y) = \frac{\frac{sin(x) cos(y)}{cos(x) cos(y)} - \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(y)}{cos(x) cos(y)}}$

And simplifying we got:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

And this identity is satisfied for all:

$(x-y) \neq \frac{\pi}{2} +n\pi$

8 0

3 years ago

An insurance company sells automobile liability and collision insurance. Let X denote the percentage of liability policies that

qaws [65]

Answer:

-764.28

Step-by-step explanation:

Given the joint cumulative distribution of X and Y as

$F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100$

#First find $F_x$ and probability distribution function , $f_x(x)$ :

$F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}$

#Have determined the probability distribution unction , $f_x(x)$ , we calculate the Expectation of the random variable X:

$E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200} dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33\\\\$

#We then calculate $E(X^2)$ :

$E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28$

Hence, the Var(X) is 764.28

4 0

3 years ago

9y+y combine the like terms to create an equivalent expression

larisa [96]

Answer:

10y

Step-by-step explanation:

9y is equal to y+y+y+y+y+y+y+y+y.

If you add y to that you get (y+y+y+y+y+y+y+y+y)+y, which is equal to 10y.

6 0

3 years ago

Read 2 more answers