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3 years ago

What is the amount of sales tax owned on a $6 book if the tax rate is 5%?

Mathematics

1 answer:

hodyreva [135]3 years ago

5 0

6/100 = 0.06
0.06 x 5 = 0.3
$0.30

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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

Can someone help me

sergij07 [2.7K]

Answer:

1) (c)∠ G = 34°

2) (a) x = 33°

3) (d) The perpendicular segments are JM, KL, PQ AND NR.

Step-by-step explanation:

Here in the given questions:

1) p II r

⇒ ∠ G = 34° (ALTERNATE EXTERIOR ANGLES)

Hence, the measure of ∠ G = 34°

2) In the given triangle:

The measure of exterior angle is always the sum of interior opposite angles.

⇒ Here, x + 72° = 105°

or, x = 105° - 72° = 33° , or x = 33°

3) The given plane in the cube is JKPN

The segments which have J , K , P and N as their one vertex are perpendicular to the given plane.

Hence, the segments are JM, KL, PQ AND NR.

7 0

4 years ago

Farmer bob has a lot of chickens and pigs. If the pigs and chickens had the total of 17 heads and 46 legs, how much of each anim

valkas [14]

11 chicken and 6 pigs is the answer

2 x 11 + 4 x 6 = 46 legs

11 + 6 = 17 heads

4 0

3 years ago

Does anyone know how to factor y^2+49?

Delvig [45]

Answer:

(y + 7i)(y - 7i)

Step-by-step explanation:

It cannot be factored using real numbers, but consider

7i × - 7i

= -49i² and i² = - 1

= 49

The factoring as a difference of squares to obtain

y² + 49 = (y + 7i)(y - 7i)

5 0

3 years ago

Read 2 more answers

Here’s a graph of a linear function. Write the equation that describes that function. (Zoom in if blurry)

Maru [420]

Answer: y = x/2 + 3

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m = change in value of y on the vertical axis / change in value of x on the horizontal axis

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

Looking at the graph,

y2 = 6

y1 = 4

x2 = 6

x1 = 2

Slope,m = (6 - 4)/(6 - 2) = 2/4 = 1/2

To determine the intercept, we would substitute x = 2, y = 4 and m= 1/2 into y = mx + c. It becomes

4 = 1/2 × 2 + c

4 = 1 + c

c = 4 - 1

c = 3

The equation becomes

y = x/2 + 3

5 0

3 years ago