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Nesterboy [21]
2 years ago
7

Help find the inverse of each of the given functions

Mathematics
1 answer:
Olenka [21]2 years ago
5 0

Answer:

x=4y-12

x+12=4y

x/4+12/4=y

y=x/4+3

therefore f^-1(x)=1/4x+3

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What is the mean median and mode of 99, 69, 96, 69, 78. Round to the nearest tenth if necessary
KiRa [710]
A median is the middle number of a data set.
To find it put the numbers in order.
78, 69, 69, 96, 99
Then find the middle number
78, 69, *69,* 96, 99

The median is 69

A mode is the number that appears the most in a set of numbers.
78, *69, 69,* 96, 99

The mode is 69

To find the mean or average of a set of numbers, add all the numbers together then divide by the number of numbers there are.
99+69+96+69+78 = 411
411/5 = 82.2

The mean is 82.2

Hope this helped! If you have anymore questions or don't understand, please comment on my profile or DM me. :)
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Which statement best describes the equation y = 3 - 4x?
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Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30
Dafna11 [192]
Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

\bf u=&#10;\begin{cases}&#10;x=7cos(330^o)\\&#10;\qquad 7\cdot \frac{\sqrt{3}}{2}\\&#10;\qquad \frac{7\sqrt{3}}{2}\\&#10;y=7sin(330^o)\\&#10;\qquad 7\cdot -\frac{1}{2}\\&#10;\qquad -\frac{7}{2}&#10;\end{cases}\qquad \qquad v=&#10;\begin{cases}&#10;x=8cos(30^o)\\&#10;\qquad 8\cdot \frac{\sqrt{3}}{2}\\&#10;\qquad \frac{8\sqrt{3}}{2}\\&#10;y=8sin(30^o)\\&#10;\qquad 8\cdot \frac{1}{2}\\&#10;\qquad 4&#10;\end{cases}

\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)&#10;\\\\\\&#10;\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~  \stackrel{b}{\frac{1}{2}}\right)\\\\&#10;-------------------------------

\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}&#10;\\\\\\&#10;\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o
8 0
3 years ago
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