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  1. Asked: April 13, 2021In: ,

    find the midpoint

    acedstud
    acedstud

    The midpoint of two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$. Therefore, the midpoint of points (-4, -2) and  (3, 3) is $\left(\dfrac{-4+3}{2}, \dfrac{-2+3}{2}\right)=\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$.

    The midpoint of two points (x_1, y_1) and (x_2, y_2) is given by \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right).

    Therefore, the midpoint of points (-4, -2) and  (3, 3) is \left(\dfrac{-4+3}{2}, \dfrac{-2+3}{2}\right)=\left(-\dfrac{1}{2}, \dfrac{1}{2}\right).

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  2. Asked: April 7, 2021In: ,

    Question of the day

    acedstud
    acedstud

    If $\tan\theta+\cot\theta=5\Rightarrow\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=5$. $\Rightarrow\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=5\Rightarrow\dfrac{1}{\sin\theta\cos\theta}=5$. $\Rightarrow\dfrac{2}{2\sin\theta\cos\theta}=5\Rightarrow\dfrac{2}{\sin2\theta}=Read more

    If \tan\theta+\cot\theta=5\Rightarrow\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=5.

    \Rightarrow\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=5\Rightarrow\dfrac{1}{\sin\theta\cos\theta}=5.

    \Rightarrow\dfrac{2}{2\sin\theta\cos\theta}=5\Rightarrow\dfrac{2}{\sin2\theta}=5\Rightarrow\sin2\theta=\dfrac{2}{5}.

    \Rightarrow2\theta=\sin^{-1}\left(\dfrac{2}{5}\right)\approx23.58^o.

    \Rightarrow\theta\approx\dfrac{23.58^o}{2}\approx11.79^o.

    Therefore, \tan2\theta+\cot\theta=\tan{23.58^o}+\cot{11.79^o}\approx5.227.

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  3. Asked: April 8, 2021In: ,

    How do I find the 6 trigonometric ratios?

    acedstud
    acedstud

    The three basic trigonometric functions are sine, cosine, and tangent. The other three trigonometric functions are cosecant, secant, and cotangent, these are the reciprocals of the three basic trigonometric functions. The easiest way to evaluate the six trigonometric functions is using a right trianRead more

    The three basic trigonometric functions are sine, cosine, and tangent. The other three trigonometric functions are cosecant, secant, and cotangent, these are the reciprocals of the three basic trigonometric functions.

    The easiest way to evaluate the six trigonometric functions is using a right triangle. Given a right triangle with an interior angle of the triangle, \theta\ne90^o, and the legs opposite and adjacent the angle \theta, we have \sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}, \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}, \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}.

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  4. Asked: April 11, 2021In: ,

    Ratio of number of boys to girls

    acedstud
    acedstud

    The ratio of boys to girls is given by $\text{number of boys} : \text{number of girls}=30:12$. Dividing through by 6, which is the common factor, we have $\dfrac{30}{6}:\dfrac{12}{6}=5:2$ Therefore, the ratio of boys to girls is 5:2.

    The ratio of boys to girls is given by \text{number of boys} : \text{number of girls}=30:12.

    Dividing through by 6, which is the common factor, we have \dfrac{30}{6}:\dfrac{12}{6}=5:2

    Therefore, the ratio of boys to girls is 5:2.

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  5. Asked: April 9, 2021In: ,

    Inverse Trigonometric Function (Problem Solving)

    acedstud
    acedstud

    Let $\theta$ be the angle of elevation of the window from the camera, then using the right triangle trigonometric ratios, $\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{100}{250}=0.4$. Thus, $\theta=\tan^{-1}0.4\approx21.8^o$. Therefore, the required angle of elevation is about 21.8 degRead more

    Let \theta be the angle of elevation of the window from the camera, then using the right triangle trigonometric ratios, \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{100}{250}=0.4.

    Thus, \theta=\tan^{-1}0.4\approx21.8^o.

    Therefore, the required angle of elevation is about 21.8 degrees.

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  6. Asked: April 9, 2021In: ,

    Polar Coordinate System

    acedstud
    acedstud

    Let the distance of James' residence from the hospital be $r$, then $x=r\cos{\theta}=50$ km and $y=r\sin{\theta}=30$ km. $r=\sqrt{x^2+y^2}=\sqrt{50^2+30^2}=\sqrt{2500+900}=\sqrt{3400}\approx{58.3}$ km. $\theta=\tan^{-1}\dfrac{y}{x}=\tan^{-1}\dfrac{30}{50}=\tan^{-1}0.6\approx31^o$. Therefore, the disRead more

    Let the distance of James’ residence from the hospital be r, then x=r\cos{\theta}=50 km and y=r\sin{\theta}=30 km.

    r=\sqrt{x^2+y^2}=\sqrt{50^2+30^2}=\sqrt{2500+900}=\sqrt{3400}\approx{58.3} km.

    \theta=\tan^{-1}\dfrac{y}{x}=\tan^{-1}\dfrac{30}{50}=\tan^{-1}0.6\approx31^o.

    Therefore, the distance of James’ residence from the hospital is about 58.3 km and the direction is 90^o+(90^o-31^0)=90^o+59^o=149^o counter clockwise.

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  7. Asked: April 11, 2021In:

    If the sum of three times of y is equal to 9, then what is the value of y?

    acedstud
    acedstud

    “The sum of three times of  is equal to 9″ means . Thus, . Dividing both sides by 3 gives . Therefore, the value of  is 3.

    “The sum of three times of y is equal to 9″ means y+y+y=9.

    Thus, 3y=9.

    Dividing both sides by 3 gives \dfrac{3y}{3}=\dfrac{9}{3}\Rightarrow y=3.

    Therefore, the value of y is 3.

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  8. Asked: April 11, 2021In: ,

    Bag containing red, green and blue balls

    acedstud
    acedstud

    The probability that none of the balls drawn is blue can be approached in many ways. I will show you two ways you can solve this problem. Method 1 Let $X$ be a random variable representing "picking a blue ball" in a random pick from 2 red, 3 green, and 2 blue balls. Then, using binomial distributionRead more

    The probability that none of the balls drawn is blue can be approached in many ways. I will show you two ways you can solve this problem.

    Method 1

    Let X be a random variable representing “picking a blue ball” in a random pick from 2 red, 3 green, and 2 blue balls. Then, using binomial distribution, p=\dfrac{2}{2+3+2}=\dfrac{2}{7}. The probability that X is equal to a value x is given by P(X=x)=^nC_xp^x(1-p)^{n-x}.

    Here, there are two draws of a ball, so n=2. Thus, the probability that none of the balls drawn is blue (x=0), is given by P(X=0)=^2C_0\left(\dfrac{2}{7}\right)^0\left(1-\dfrac{2}{7}\right)^{2-0}

    =1\times1\times\left(\dfrac{5}{7}\right)^2=\dfrac{25}{49}\approx0.5102.

     

    Method 2

    The probability that ‘the first’ ball is not blue is 1 minus the probability of blue = 1-\dfrac{2}{7}=\dfrac{5}{7}.

    The probability that ‘the second’ ball is not blue is 1 minus the probability of blue = 1-\dfrac{2}{7}=\dfrac{5}{7}.

    The probability that ‘the first’ ball is not blue AND ‘the second’ ball is not blue  = \dfrac{5}{7}\times\dfrac{5}{7}=\dfrac{25}{49}\approx0.5102.

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  9. Asked: April 9, 2021In: ,

    Rational function

    acedstud
    acedstud

    The $x$-intercepts of a function, $f(x)$, are the values of $x$ when $f(x)=0$. Given $f(x)=\dfrac{x^2 -15}{x^2-2x-3}$. Letting $f(x)=0$, we have $\dfrac{x^2 -15}{x^2-2x-3}=0\Rightarrow x^2-15=0$        {$\because$ multiplying both sides by the denominator.] Adding 15 to both sides and then taking thRead more

    The x-intercepts of a function, f(x), are the values of x when f(x)=0.

    Given f(x)=\dfrac{x^2 -15}{x^2-2x-3}.

    Letting f(x)=0, we have \dfrac{x^2 -15}{x^2-2x-3}=0\Rightarrow x^2-15=0        {\because multiplying both sides by the denominator.]

    Adding 15 to both sides and then taking the square root of both sides, we have x^2=15\Rightarrow x=\pm\sqrt{15}.

    Therefore, the x-intercepets of the function are at x=\pm\sqrt{15}.

     

    The y intercepts of a function are the values of f(x) when x=0. Thus, equating x=0 in the function, we have f(0)=\dfrac{0^2 -15}{0^2-2(0)-3}=\dfrac{-15}{-3}=5.

    Therefore, the y intercept is at y=5.

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  10. Asked: April 10, 2021In: ,

    Simplify the following

    acedstud
    acedstud

    Some of your questions are not clear enough, however, I will give you the answers to the extent I can understand what you are asking. If I did not interpret any of your questions correctly, you can ask them again. Note that any number raised to an exponent of 0 is 1. 1.) $-2^3-3^0=-(2\times2\times2)Read more

    Some of your questions are not clear enough, however, I will give you the answers to the extent I can understand what you are asking. If I did not interpret any of your questions correctly, you can ask them again.

    Note that any number raised to an exponent of 0 is 1.

    1.) -2^3-3^0=-(2\times2\times2)-1=-8-1=-9.

    2.) -1^{-5} y²=-\dfrac{1}{1^5}y^2=-\dfrac{1}{1\times1\times1\times1\times1}y^2=-\dfrac{1}{1}y^2=-y^2

    3.) (2^{-2})^2=2^{-2\times2}=2^-4=\dfrac{1}{2^4}=\dfrac{1}{2\times2\times2\times2}=\dfrac{1}{16}.

    4.) I don’t understand your question here. Can you reask  the question or attach the question in a fresh question.

    5.) -4^0b^{-4}=-1\cdot\dfrac{1}{b^4}=-\dfrac{1}{b^4}.

     

    PS. Not that there is a difference between -a^x and (-a)^x. So, be more explicit when asking questions.

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