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miskamm [114]
2 years ago
15

What is the solution to the linear equation?

Mathematics
1 answer:
tangare [24]2 years ago
4 0

2.3x - 5.6x = 1.3 + 1.2 \\  - 3.3x = 2.5 \\  - x = 0.75 \\ x =  - 0.75

i hope it helps u

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Write the equation for the inverse of the function.<br> y = sin x<br>​
MakcuM [25]

Answer:

<em>  The answer is \sin^{-1}x = y</em>

Step-by-step explanation:

    We need to find the inverse of the function

     y = sinx

    Taking \sin^{-1} on both sides

    \sin^{-1}y =  \sin^{-1}sinx

     \sin^{-1}y = x

     Substituting x to y and y to x

<em>      \sin^{-1}x = y</em>

     This is the required inverse of the function

     The graph of the function and its inverse is symmetric about y = x

<em>     The answer is \sin^{-1}x = y</em>

7 0
2 years ago
Read 2 more answers
What is the volume of the given prism? Round the answer to the nearest tenth.
kicyunya [14]
\bold{ANSWER:}
OPTION C

\bold{SOLUTION:}

6 0
2 years ago
Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val
zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill

\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}

\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}

6 0
2 years ago
Anyone know these two questions??
gogolik [260]

1):  Answer  =====>  Letter Choice, (B),  The rate in which you walked home, The slope is the same, as the Rate of Change, In Distance Divided by the Change in Time  =====>    d    =     4   -1/15t(4) / 1/15t

 

2):   The Equation:  d     =       -1/15t     +     4

The more the Number of Minutes:  Larger the Negative Quantity;  ====>   -1/15t

The Larger the Negative Quantity in Equation:  D value is Smaller:

Therefore your Answer ===>  The Distance from home (d), Decreased with Minute Walked (t)  ====>  So, Letter Choice,  (C)



2):   Towards your home, The slope is Negative, and the Distance (y - value), Decreases, as the Time  (x- Values),  Increases. Notice the Slope is Negative, Because, you are getting closer to home.






Hope that helps!!!!                                           : )

7 0
2 years ago
5x/p = w<br> solve for x
trasher [3.6K]

Answer:

x=wp/5

Step-by-step explanation:

7 0
2 years ago
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