A/g=1 solve for a how do I get my answee to this problem

Mathematics

1 answer:

Mama L [17]2 years ago

3 0

Answer:

a=g

Step-by-step explanation:

If you are solving for 'a', then you just need to get 'a' separated on its own. To do this, you can multiply both sides of the equation by 'g'. Now you have "a=1g" which simplifies into "a=g"

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How to do this question plz

hjlf

Answer:

$\underline{ \boxed{ \underline{part \: a}}}. \\the \: required \: distance \: a part \: be \to \: \boxed{\underline {\overline{p } =79.23 \: km} } \\ \\ \underline{ \boxed{ \underline{part \: b}}}. \\ the \: bearing \: of \: \boxed{ B } \: \: from \: \boxed{ \underline{A}} \: is \to \boxed{ \underline{324 }\degree }$

Step-by-step explanation:

$\underline{ \boxed{ \underline{part \: a}}}. \\ l et \: the \: required \: distance \: apart \: be \to \: \boxed{ \overline {p}} \\ let \: \boxed{ \angle \: P} \: be \: the \: angle \: of \: seperation \: between \: both \: ships :given \: by \to \\ P = (244 - 180) - (196 - 180 )= (64 -16) = \boxed{48 \degree} \\ if \: ship \: A \: and \: B \: left \: the \: port\: at \: (10:30) \: \\ then \: at \: (14 :00) \: their \: time \: interval \: would \: be \to \\ (14 : 00) - (10 : 30) = \boxed{ 3.5 \: hrs} \\ ship \: \boxed{A }\: distance \: from \: the \: port =( v \times t) = (30 \times 3.5) = \boxed{105 \: km} \\ ship \: \boxed{B }\: distance \: from \: the \: port =( v \times t) = (24 \times 3.5) = \boxed{84\: km} \\n ow........we \: cant \: do \: much \: if \: their \: are \: no \: angles \: to \: work \: with \to \\ applying \:t he \: cosine \: rule : we \: have \to \\ \cos(P) = \frac{ {a}^{2} + {b}^{2} - \overline{{p}^{2} } }{2(ab)} \\ {a}^{2} + {b}^{2} - \overline{{p}^{2} } = 2(ab) \cos(P) \\ \overline{{p}^{2} } = {a}^{2} + {b}^{2} - \{2(ab) \cos(P) \} \\ \overline{{p}^{2} } = {84}^{2} + {105}^{2} - \{2(84)(105) \cos(48 \degree) \} \\\overline{{p}^{2} } = 18,081 - 11,803.463897 \\ \overline{{p}^{2} } = 6,277.536103 \\ \overline{p } = \sqrt{6,277.536103} \\ \boxed{\underline {\overline{p } =79.23 \: km} } \\ \\ \underline{ \boxed{ \underline{part \: b}}}. \\ to \: sove \: this : we \: first \: find \: angle \: \angle \: B : by \: applying \to \\ \frac{ \sin(B) }{b} = \frac{ \sin(P) }{p} \\ p \ast \sin(B) = b \ast \sin(P) \\ B = \sin {}^{ - 1} ( \frac{ b \ast \sin(P)}{p} ) \\ B = \sin {}^{ - 1} ( \frac{ 105 \times \sin(48 \degree)}{79.230903712} ) \\ \\ B = \sin {}^{ - 1} ( \frac{ 78.030206678}{79.230903712} ) \\ \\ B = \sin {}^{ - 1} (0.9848455971) \\ \boxed{B = 80 \degree} \\ hence \ : the \: bearing \: of \: \boxed{ B } \: \: from \: \boxed{ \underline{A}} \: is \to \\ 360 - \{180 - (80 + 48 + 16) \} \\ 360 -(180 - 144 )\\ 360 -36 = \boxed{324 \degree }$

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