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Tamiku [17]
3 years ago
6

What is the reciprocal of -11/30

Mathematics
1 answer:
Burka [1]3 years ago
3 0
The answer is -0.00303030303
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Below are five number sequences.
oksian1 [2.3K]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
5/8
irga5000 [103]

Answer:

5/12

Step-by-step explanation:

Total ear buds in Mrs. Jackson’s class = 5/8

Number of black ear buds in Mrs. Jackson’s class = 2/3

What fraction of

the scholars have black ear buds?

2/3 of the scholars have black ear buds in Mrs. Jackson’s class

2/3 of 5/8

= 2/3 × 5/8

= 10/24

= 5/12

Fraction of the scholars that have black ear buds = 5/12

3 0
2 years ago
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
help?‎‎‎‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎
vivado [14]

Answer:

Step-by-step explanation:

(-6,-5) , ( 0,-1)  , (-3 , -3) , (2 , 0 ) are solution of the inequalities.

The point that lies on the purple area and on the solid line are the solutions of the equations.

5 0
2 years ago
In the first quadrant you start at (7,3) and move 5 unit left
AveGali [126]

Answer:

That would be (2,3)

Step-by-step explanation:

When we're moving left in the coordinate plane, we are subtracting the x-coordinate by how much distance we move.

In this case, 7-5=2, so the new x-coordinate is 2, but the y-coordinate doesn't change since we're going horizontally and not vertically.

7 0
3 years ago
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