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gulaghasi [49]
3 years ago
5

Let f(x) = cos(x). Find the x-intercepts of f(x) on [0, 2π).

Mathematics
2 answers:
JulsSmile [24]3 years ago
6 0

the x-intercepts of cosx are

\frac{\pi }{2} and \frac{3\pi }{2} on [0, 2π )


Talja [164]3 years ago
4 0

Answer:

Tthe x-intercepts of f(x)= cosx on [0, 2π ) are:

\frac{\pi }{2} and \frac{3\pi }{2}

Step-by-step explanation:

x-intercept are those values of x for which f(x)=0

So, x-intercept of  f(x)=cosx will be those values of x for which

cosx=0

cosx=0 for  

\frac{\pi }{2} and \frac{3\pi }{2} on [0, 2π )

Hence, the x-intercepts of f(x)= cosx on [0, 2π ) are:

\frac{\pi }{2} and \frac{3\pi }{2}

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1 1/8 +5/6 =<br><br> And then subtract 1/2 from the answer
kari74 [83]
Alright so start of with solving 1 1/8+5/6

Solve and you get 1 23/24

Now to subtract you have your answer subtracted by 1/2:

1 23/24-1/2=1 11/24

And so you get 1 11/24 

~Evie


8 0
3 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
Three friends Kofi,Yaw and Musa are in partnership business in which Kofi and Yaw contributed 5/9 and 2/15 of the cost of 14000.
wlad13 [49]

Answer:

(a) Total = 45000

(b) Yaw, Musa and Kofi

Step-by-step explanation:

Given

Kofi =5/9

Yaw = 2/15

Musa = 14000

Solving (a): The cost of the business

First, we solve for the fraction of Musa

Musa + Kofi + Yaw = 1

Musa = 1 - (Kofi + Yaw )

Musa = 1 - (5/9 + 2/15)

Musa = 1 - (\frac{25+6}{45})

Musa = 1 - \frac{31}{45}

Musa = \frac{45 - 31}{45}

Musa = \frac{14}{45}

So, we have:

\frac{14}{45} * Total = 14000

Make Total the subject

Total = 14000 * \frac{45}{14}

Total = 1000 * 45

Total = 45000

Solving (b): Partners in ascending order

First, represent the fractions as decimals

Kofi =5/9 = 0.5556

Yaw = 2/15 = 0.1333

Musa = \frac{14}{45} = 0.3111

From the conversion above, the least is 0.1333 (Yaw), then 0.3111 (Musa), then 0.5556 (Kofi).

<em>So, the order is: Yaw, Musa and Kofi</em>

6 0
3 years ago
When a number cannot be divided evenly, the amount left over is called the
scoundrel [369]

Answer:

The remainder.

Step-by-step explanation:

I hope this helps.

8 0
3 years ago
In simplest radical form, what are the solutions to the quadratic equation 0 = –3x2 – 4x + 5?
JulsSmile [24]
- 3 x² - 4 x + 5 = 0
x_{1} = \frac{4+ \sqrt{16+60} }{-6} =  \frac{4+8.718}{-6}
x1 = -2.12
x_{2} = \frac{4-8.718}{-6}
x2= 0.786

4 0
3 years ago
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