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2 years ago

What is the circumference and the area of a circle with diameter 5 yd?

Mathematics

1 answer:

valina [46]2 years ago

5 0

The circumference is pi times 5.
So the circumference is 15.7

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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 $\sf \theta$

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

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

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

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

Solving the given expression :- 

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

Putting,

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

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

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

Using (a + b ) (a - b ) = a² - b²

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

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

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

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

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

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

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

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

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

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

✧ 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

1 year ago

At a bake sale for every 8 cupcakes sold you earn 5.00. If u made 45 how many cupcakes did you sell?

grandymaker [24]

45 divided by 5 is 9 and 9 times 8 is 72 so the answer is 72.

3 0

2 years ago

Oliver interviewed 30% of the 9th grade class and 70% of the 10th grade class at his school. Jenny interviewed 75% of the 9th gr

julia-pushkina [17]

Answer:

A. 36

Step-by-step explanation:

We are given a total of 176 interviewed by Oliver and a total of 140 interviewed by Jenny. To find how many more 10th graders than 9th graders were interviewed, subtract the totals given

176 - 140 = 36

This is how we came to the answer:

We are given 70% of the 10th-grade and 30% of the 9th-grade with a total of 176 for Oliver.

While we're given 75% of the 9th-grade class and 25% of the 10th-grade with a total of 140 interviewed by Jenny

Oliver's Interviewees

10-graders

Firstly, let's find what the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

70% of 176 =

Cross multiply

123.2 were 10-graders interviewed by Oliver

9th-graders

Now, to find the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

30% of 176 =

Cross multiply

52.8 were 9th-graders interviewed by Oliver

Jenny's Interviewees

9th-graders

Firstly, let's find what the number of 9th-graders was interviewed by Jenney; find the percentage of the 9th-graders by the total;

75% of 140 =

Cross multiply

105 students were 9th-graders interviewed by Jenney.

10th-graders

Now, to find the number of 10th-graders was interviewed by Jenney; find the percentage of the 10th-graders by the total;

25% of 140 =

Cross multiply

35 students were 10th-graders interviewed by Jenney.

Total calculation

Use the results and sum them up by 9th-grade plus 9th-grade and 10th-grade plus 10-grade. Then subtract the amount gotten from 9th-grade away from the amount gotten from 10th-grade;

Oliver's 9th-grade = 52.8

Jenny's 9th-grade = 105

105 + 52.8 = 157.8

Oliver's 10th-grade = 123.2

Jenny's 10th-grade = 35

123.2 + 35 = 158.2

Total calculation: 158. 2 - 157.8 = 0.4

Therefore, there are 36 more 10th than 9th.

For more information, visit: brainly.com/question/23490909

7 0

2 years ago

Use decimals and the data for 1988 and 1996 to find the slope intercept form of a trend line

Tju [1.3M]

The answer would be y = 312.5x + 1564

hope this helped :)
alisa202

4 0

2 years ago

A vacuum had been marked down by 20% and solf for $250. What was the original price of the vacuum?

igomit [66]

Answer:

The original price was $312.50

Step-by-step explanation:

If the vacuum was marked down by 20%, that means it was sold for 80% of the full price. It was sold for $250, so 80% of the original price is $250.

Let's set up an equation.

80/100 of x = 250

0.8 • x = 250

Divide both sides by 0.8.

x = 312.50

The original price was $312.50

Hope this helps!

7 0

2 years ago

Read 2 more answers