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kow [346]
3 years ago
8

Make x the subject of the formula х +12c = 5d 63

Mathematics
1 answer:
dezoksy [38]3 years ago
7 0
X=5db^3-2cb^3
Workup in photo below.
Good luck

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Please answer this question, i request
Jet001 [13]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

\star  \:  \tt \cot  \theta = \dfrac{7}{8}

{\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}

\star \:  \tt \dfrac{(1  +  \sin \theta)(1 - \sin \theta) }{(1 +  \cos \theta) (1  -  \cos \theta) }

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Consider a \triangle ABC right angled at C and \sf \angle \: B = \theta

Then,

‣ Base [B] = BC

‣ Perpendicular [P] = AC

‣ Hypotenuse [H] = AB

\therefore \tt \cot  \theta   =  \dfrac{Base}{ Perpendicular}  =  \dfrac{BC}{AC} = \dfrac{7}{8}

Let,

Base = 7k and Perpendicular = 8k, where k is any positive integer

In \triangle ABC, H² = B² + P² by Pythagoras theorem

\longrightarrow \tt {AB}^{2}  =   {BC}^{2}  +   {AC}^{2}

\longrightarrow \tt {AB}^{2}  =   {(7k)}^{2}  +   {(8k)}^{2}

\longrightarrow \tt {AB}^{2}  =   49{k}^{2}  +   64{k}^{2}

\longrightarrow \tt {AB}^{2}  =   113{k}^{2}

\longrightarrow \tt AB  =   \sqrt{113  {k}^{2} }

\longrightarrow \tt AB = \red{  \sqrt{113}  \:  k}

Calculating Sin \sf \theta

\longrightarrow  \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}

\longrightarrow  \tt \sin \theta = \dfrac{AC}{AB}

\longrightarrow  \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }

\longrightarrow  \tt \sin \theta =  \purple{  \dfrac{8}{ \sqrt{113} } }

Calculating Cos \sf \theta

\longrightarrow  \tt \cos \theta = \dfrac{Base}{Hypotenuse}

\longrightarrow  \tt \cos \theta =  \dfrac{BC}{ AB}

\longrightarrow  \tt \cos \theta =  \dfrac{7 \cancel{k}}{ \sqrt{113} \:  \cancel{k } }

\longrightarrow  \tt \cos \theta =  \purple{ \dfrac{7}{ \sqrt{113} } }

<u>Solving the given expression</u><u> </u><u>:</u><u>-</u><u> </u>

\longrightarrow \:  \tt \dfrac{(1  +  \sin \theta)(1 - \sin \theta) }{(1 +  \cos \theta) (1  -  \cos \theta) }

Putting,

• Sin \sf \theta = \dfrac{8}{ \sqrt{113} }

• Cos \sf \theta = \dfrac{7}{ \sqrt{113} }

\longrightarrow \:  \tt \dfrac{ \bigg(1 +  \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 +  \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)}

<u>Using</u><u> </u><u>(</u><u>a</u><u> </u><u>+</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>(</u><u>a</u><u> </u><u>-</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>a²</u><u> </u><u>-</u><u> </u><u>b²</u>

\longrightarrow \:  \tt  \dfrac{ { \bigg(1 \bigg)}^{2}  -  { \bigg(  \dfrac{8}{ \sqrt{133} } \bigg)}^{2}   }{ { \bigg(1 \bigg)}^{2}  -  { \bigg(  \dfrac{7}{ \sqrt{133} } \bigg)}^{2}  }

\longrightarrow \:  \tt   \dfrac{1 -  \dfrac{64}{113} }{ 1 - \dfrac{49}{113} }

\longrightarrow \:  \tt   \dfrac{ \dfrac{113 - 64}{113} }{  \dfrac{113 - 49}{113} }

\longrightarrow \:  \tt { \dfrac  { \dfrac{49}{113} }{  \dfrac{64}{113} } }

\longrightarrow \:  \tt   { \dfrac{49}{113} }÷{  \dfrac{64}{113} }

\longrightarrow \:  \tt    \dfrac{49}{ \cancel{113}} \times     \dfrac{ \cancel{113}}{64}

\longrightarrow \:  \tt   \dfrac{49}{64}

\qquad  \:  \therefore  \:  \tt \dfrac{(1  +  \sin \theta)(1 - \sin \theta) }{(1 +  \cos \theta) (1  -  \cos \theta) }  =   \pink{\dfrac{49}{64} }

\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered}

{\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}

✧ Basic Formulas of Trigonometry is given by :-

\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \:  \sf{ In \:a \:Right \:Angled \: Triangle :}  \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

✧ Figure in attachment

\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered}

3 0
2 years ago
Solve for x <br> (6x+5) (6x+16) (7x-3)
ololo11 [35]

Answer:

i think it is 43

Step-by-step explanation:

5 0
3 years ago
A. 4 packs of pencils and 3 packs of erasers B. 4 packs of pencils and 5 packs of erasers C. 5 packs of pencils and 3 packs of e
amid [387]
C. 5 packs of pencils and 3 packs of erasers. it is c because no matter what 20(the pack of erasers) is multiplied by it will always end in a zero, therefore you would have to multiply 12 by a number that would make it a multiple of 20 as well which happens to be 5. 5×12=60 and 3×20=60
3 0
3 years ago
Is this relation a function? EXPLAIN WHY OR WHY NOT<br> {(-3,2), (-4, 4), (-3, 3), (4,4)}
denis-greek [22]
No it is not a function, because a function cannot have more than one output per input. When x=-3, there are two solutions shown: (-3,2) and (-3,3), therefore it cannot be a function.
3 0
2 years ago
You are choosing between two health clubs. Club A offers membership for a fee of $ 19plus a monthly fee of $ 21.  Club B off
11111nata11111 [884]

Answer:

For 2 months

Step-by-step explanation:

Let after x months the cost of each health club is same,

Now, In club A,

Membership fees = $ 19,

Monthly fees = $ 21,

So, the total fees for x months = membership fees + total monthly fees for x months

= 19 + 21x

In Club B,

Membership fees = $ 23,

Monthly fees = $ 20,

So, the total fees for x months = membership fees + total monthly fees for x months

= 23 + 20x

Thus, we can write,

19 + 21x = 23 + 20x

21x - 20x = 23 - 21

x = 2

Hence, for 2 months the total cost of each health club would be same.

8 0
3 years ago
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