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3 years ago

Which function in the inverse of f(x) = 2x+3

Mathematics

1 answer:

mixer [17]3 years ago

3 0

Answer: if im right it might be x=-3 although im not postive

Step-by-step explanation:

if you make the equation x=2x+3 you minus 2x on to both sides giving you -x=3 so than you make x positive and 3 negative (if im correct)

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telo118 [61]

Answer:

Step-by-step explanation:

7 0

2 years ago

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How to write a real world situation that could be modeled by the equation 100-6x=160-10x

blondinia [14]

100-6x=160-10x
100 - 160 = -10x +6x
-4x = -60
x = -60 / -4
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5 0

3 years ago

Which statements are true about the median of a data set?

Leto [7]

Answer: Third and fifth option

3) The median is the number in the middle on an ordered set of data.

5) To find the middle of an even data set, find the average of the two middle numbers.

Step-by-step explanation:

Given a set of data ordered from lowest to highest, the median of the data is the number that is in the middle. In other words, it is a number x for which it is true that 50% of the data is greater than x and the other 50% of the data is less than x.

For example, for the following set of 7 data:

2, 3, [5], 8, 13

the median is 5.

If the data number is even, for example 10 data:

3, 5, 9, 12, [19, 21], 33, 35, 40, 69

The median is the average between the two data that are in the middle

$\frac{19 +21}{2} = \frac{40}{2} = 20$

Note that the median only represents a partition of the ordered data set. Therefore, it is not affected by outlier. For example

2, 4, 4.5, 4.8, 5 Median = 4.5

2, 4, 4.5, 67, 1506 Median = 4.5

The median does not represent the difference between the highest and the lowest data.

Therefore the correct affirmations are:

3) The median is the number in the middle on an ordered set of data.

5) To find the middle of an even data set, find the average of the two middle numbers.

8 0

3 years ago

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What are the factors of six

Varvara68 [4.7K]

6 12 18 24 30 36 42 48

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3 years ago

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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