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sammy [17]
2 years ago
12

Please help me bro please

Mathematics
2 answers:
GaryK [48]2 years ago
6 0

Answer:

  • n = 11°

⠀

Step-by-step explanation:

  • When two straight lines intersect each other, then the pairs of angles so formed without any common arm are called vertically opposite angles.

  • Vertically opposite angles are equal to each other.

⠀

So,

{\longrightarrow \it\qquad { \ { (6n   - 4) {}^{ \circ}   =  {(5n + 7)}^{ \circ} }}}

⠀

Removing the brackets,

{\longrightarrow \it\qquad { \ { 6n   - 4{}^{ \circ}   =  {5n + 7 \: }^{ \circ} }}}

{\longrightarrow \it\qquad { \ { 6n   - 5n   =  {  7 \: }^{ \circ} +4{}^{ \circ} }}}

{\longrightarrow \it\qquad { \pmb{ n   =  {  11 \: }^{ \circ}  }}}

⠀

Therefore,

  • The value of n is 11°
Marianna [84]2 years ago
4 0

We will use the <em>concept</em> of Vertically Opposite angle here, i.e at the <em>point</em> of Intersection of two straight lines, the VOA (Vertically Opposite Angles) are <em>equal</em>

By using this property we will be <em>having</em> ;

{:\implies \quad \sf (6n-4)^{\circ}=(5n+7)^{\circ}}

{:\implies \quad \sf 6n-5n=7+4}

{:\implies \quad \bf \therefore \quad \underline{\underline{n=11}}}

<em>This is the required answer </em>

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Check the picture below.

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\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

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