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mash [69]
3 years ago
6

Ms. Dyna a mathematics teacher has a salary of 120 000$ per month, if salary increases by 9% calculate the amount of extra she r

eceives a month
Mathematics
1 answer:
Studentka2010 [4]3 years ago
8 0

Answer:

129,000

Step-by-step explanation:

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  ¹9.2

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A cable service provider charges $40 per month for its basic package plus an additional $5.35 for each premium channel chosen. T
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The ratio of water and milk in a mixture is 2:3, what percent of water is mixed in the mixture?
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40% water 60% milk

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if you make then go to 100 by multiplying 2:3 by 2 you get 4:6 and then add a 0 at the end and there’s your answer
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Help______________________________________
ser-zykov [4K]

The right step followed in the system to get he equations is (D). To get the system B. The second equation in system A was replaced by the sum of that equation and the first equation multiplied by 6. The solution to system B will be the same as the solution of system A.

<h3>Meaning of Simultaneous Equation.</h3>

A simultaneous equation can be defined as an equation that possesses two unknowns and one constant.

The solution to the problem can only be gotten by solving it side by side with another equation of the same format.

In conclusion, The right step followed in the system to get he equations is (D). To get the system B. The second equation in system A was replaced by the sum of that equation and the first equation multiplied by 6. The solution to system B will be the same as the solution of system A.

Learn more about simultaneous equation: brainly.com/question/148035

#SPJ1

3 0
2 years ago
Simplify the expression. tan(sin^−1 x)
Blizzard [7]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(that is the range of the inverse sine function).


So,

\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}


Square both sides:

\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}


Since \mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},} then \mathsf{cos\,\theta} is positive. So take the positive square root and you get

\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}


Then,

\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}


I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

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