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3 years ago

the hypotenuse of a right triangle is 3 cm and one of the legs is 1 cm what is the length of the other leg ?

Mathematics

1 answer:

Afina-wow [57]3 years ago

4 0

Answer:

2 $\sqrt{2}$

Step-by-step explanation:

let the other leg be x

Using Pythagoras' identity

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

x² + 1² = 3²

x² + 1 = 9 ( subtract 1 from both sides )

x² = 8 ( take the square root of both sides )

x = $\sqrt{8}$

= $\sqrt{4(2)}$

= $\sqrt{4}$ × $\sqrt{2}$ = 2 $\sqrt{2}$ ← exact value

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PLEASE HELP I AM ON A TIME LIMIT

Alinara [238K]

Answer:

(a) even: J, K, M, O; odd: L, N

(b) L and O are connected to J

Step-by-step explanation:

Count the edge ends that intersect each vertex. You get ...

J-2, K-2, L-3, M-2, N-3, O-4

These numbers are the degree of the vertex.

a) Vertices with even degree are J, K, M, O, since these have degrees of 2, 2, 2, and 4, respectively--all even numbers.

Vertices with odd degree are L and N, since these both have degree 3, an odd number.

b) The vertices that are adjacent to J are the ones at the other ends of the edges that intersect vertex J. One of those two edges connects to vertex L; the other to vertex O. J is adjacent to L and O.

c) When we counted edges in part (a), we found vertex N to be of degree 3.

5 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 $\sf \theta$

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

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

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

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

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

A spinner is divided into 8 equal sections. Two of the sections are red, 3 are green, and 3

Marina86 [1]

Answer:

part a: 1/2, two red, probability is 1/2

part b: 450, 150 times 3 =450, hope it helps

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3 years ago

What is the log 5x=log (2x+9)

Ray Of Light [21]

5x = 2x + 9, where 5x > 0 and 2x + 9 > 0 ;
Then, 3x = 9 ;
Then, x = 3 ;
Verify : 5 × 3 = 2 × 3 + 9 ;correct !
5 × 3 > 0 ; correct !
2 × 3 + 9 > 0; correct !

8 0

3 years ago

How to find the perimeter of shapes

Morgarella [4.7K]

You add the legnths of the sides together exg
if you have rectangle with width=2 legnth=3
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3 years ago