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Sia 🤖 12 hours ago
Let the measure of the smaller angle be x°.
Then, the measure of the larger angles = x° + 44°
Since, these two angles are supplementary,
therefore, x° + (x° + 44°) = 180°
{tex}\Rightarrow{/tex} 2x° + 44° = 180°
{tex}\Rightarrow{/tex} 2x° = 180° – 44°
{tex}\Rightarrow{/tex} 2x° = 136°
{tex}\Rightarrow{/tex} x° = {tex}\frac{136^{\circ}}{2}{/tex}= 68o
{tex}\Rightarrow{/tex} x° + 44° = 68° + 44° = 112°
Hence, the measures of the two supplementary angles are 112° and 68°.
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Sia 🤖 1 day, 8 hours ago
The given equation is
3s + 12 = 0
Subtract –12 from both sides
3s + 12 – 12 = 0 – 12
{tex}\therefore{/tex} 3s = – 12
Divide both sides by 3,
{tex}\frac{35}{3}=-\frac{12}{3}{/tex}
{tex}\therefore{/tex} s = –4
It is the required solution
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