Qual o valor que Z assume? -Cz+ 6z = Tz+ 83

Mathematics

1 answer:

o-na [289]3 years ago

3 0

Answer:

$\large \boxed{z = \dfrac{83}{6 - C - T}}$

Step-by-step explanation:

$\begin{array}{rccl}-Cz + 6z & = & Tz + 83 & \\6z -Cz - Tz &= & 83 & \text{Subtracted Tz from each side}\\z(6 - C - T) & = & 83 & \text{Removed the common factor}\\z& = & \mathbf{\dfrac{83}{6 - C - T}} &\text{Divided each side by 6 - C - T}\\\end{array}\\\\\large \boxed{\mathbf{z = \dfrac{83}{6 - C - T}}}$

You might be interested in

Cos(-170°) = _____<br><br> -cos10°<br> cos10°<br> cos(-10°)

ikadub [295]

<em>The right answer for:</em>

<h2>Explanation:</h2>

The cosine function is an even function, which means that for every point $(x,y)$ on the graph of $y=cos(x)$ then the point $(-x,y)$ also lies on the graph of the function. In other words, we can write:

$cos(-170^{\circ})=cos(170^{\circ})$

But:

$170^{\circ}=180^{\circ}-10^{\circ}$

So:

$cos(170^{\circ})=cos(180^{\circ}-10^{\circ})$

By property:

$cos(a-b)=cosacosb+sinasinb \\ \\ a=180^{\circ} \\ \\ b=10^{\circ} \\ \\ cos(170^{\circ} )=cos(180^{\circ} -10^{\circ})=cos180^{\circ} cos10^{\circ} +sin180^{\circ} sin10^{\circ} \\ \\ cos(170^{\circ})=-cos(10^{\circ})+0 \\ \\ \boxed{cos(170^{\circ})=-cos(10^{\circ})}$