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2 years ago

Simplify each expression and state whether each is a volume, an area, or neither.

Mathematics

1 answer:

4vir4ik [10]2 years ago

4 0

a. 2n^4b, neither a volume nor an area

b. 2Iw + 2wh + 2Ih, It is an area

c. 80 y^4, neither a volume nor an area

<h3>How to simply the algebraic expressions</h3>

a. n(4) 2 x b

n * n * n* n* 2 * b

2n^4b

It is neither an area nor a volume

b. lw+lw+wh+wh+lh+lh

Collect like terms

2Iw + 2wh + 2Ih

It is an area

c. 8y2(10y2)
8y^2 * 10 * y^2

80 y^4

It is neither a volume nor an area

Thus, the expressions are simplified as a. 2n^4b, b. 2Iw + 2wh + 2Ih, c. 80 y^4

Learn more about algebraic expressions here:

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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

2 years ago

What set of transformations is performed on triangle ABC to form triangle A′B′C′?

Simora [160]

Answer: A translation 5 units down followed by a 180-degree counterclockwise rotation about the origin .

Step-by-step explanation:

From the given figure, the coordinates of ΔABC are A(-3,4), B(-3,1), C(-2,1) and the coordinates of ΔA'B'C' are A'(3,1), B'(3,4), C'(2,4).

When, a translation of 5 units down is applied to ΔABC, the coordinates of the image will be

Then applying 180° counterclockwise rotation about the origin, the coordinates of the image will be :-

which are the coordinates of ΔA'B'C'.

Hence, the set of transformations is performed on triangle ABC to form triangle A’B’C’ is " A translation 5 units down followed by a 180-degree counterclockwise rotation about the origin ".

4 0

4 years ago

The city buses that run in your neighborhood stop every 12 minutes .The first bus arrives at 5:00 am for passengers.You get to t

Irina-Kira [14]

Answer:

Therefore you will have to wait 11 minutes.

Step-by-step explanation:

Find the number of minutes between 5:00am and 8:37am

60 mins in an hour.

8-5=3 hours

60x3=180+37=217 minutes

Divide 217 by 12 to see how many bus stops have already been made

217/12=18

18*12=216

217-216=1 minute

12-1=11 minutes

Hope this helps!

8 0

3 years ago

A tree that is next to a telephone pole is 15 feet tall and casts a shadow 4 feet long.

mestny [16]

Answer:

Step-by-step explanation:

15 =4

15x 6=90

4x6=24

5 0

3 years ago

Read 2 more answers

What is 5/8 with a denominator of 24?

Arturiano [62]

Answer:

Step-by-step explanation:

you multiply 8*3 to get 24 so you do the same to 5. so 5*3 = 15 so 15/24

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3 years ago