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3 years ago

Need help please thanks

Mathematics

2 answers:

pav-90 [236]3 years ago

8 0

This equation is in standard form, if you put it into slope-intercept form, you will be able to graph it. Once you graph it you will be able to find the intercepts.

5x - 3y = 15

-3y = -5x + 15

y = 5/3x - 5

X-Intercept: (3,0)

Y-Intercept: (0, -5)

frozen [14]3 years ago

5 0

Answer:

x-intercept = 3 ; y-intercept = -5

Step-by-step explanation:

Lets rearrange the equation into the form y = mx + b:

5x - 3y = 15

-3y = 15 - 5x

Divide by -3:

y = 5x/3 - 5

Here we know that b (-5) is the y-intercept, and we can find the x-intercept by starting at the y-intercept and using the slope till we get the x-intercept. Or you can set y = 0 and solve for x, either way you get x-intercept = 3.

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17. Farimah and Helio are standing 15 ft. apart from each other and looking up at a kite that is with the flying between them. F

miss Akunina [59]

Answer:

Helio is 53.2 feet from the kite ⇒ the last answer

Step-by-step explanation:

* Lets change this story problem to a trigonometry problem

- Assume that there is a triangle joining between the

kite, Farimah and Helio

- The name of the triangle is KFH, where K position of the kite,

F position of Farimah and H is the position of Helio

∵ Farimah and Helio are standing 15 feet apart from each other

∴ FH = 15 feet

∵ Farimah is flying the kite on a 57 feet string at an angle

of 68 with the ground

∴ FK = 57 feet

∴ m∠KFH = 68°

∵ We need to know that Helio is how far from the kite

∴ We need to calculate the length of KH

* Now lets find the best way to find the length of KH

using the trigonometry

- We have the length of two sides and the measure of the included

angle between them , then the best way is the cosine Rule

* Lets explain the cosine rule:

- In ΔABC:

∵ a is the length of the side opposite to ∠A ⇒ a is BC

∵ b is the length of the side opposite to ∠B ⇒ b is AC

∵ c is the length of the side opposite to ∠C ⇒ c = AB

∴ a² = b² + c² -2bc × cos(A)

∴ b² = a² + c² -2ac × cos(B)

∴ c² = a² + b² -2ab × cos(C)

* We will use the rule in our problem to find HK

∵ FH is k , HK is f , KF is h

∴ f² = h² + k² - 2hk × cos(F)

∵ h = 57 feet , k = 15 feet , m∠F = 68°

∴ f² = (57)² + (15)² - 2(57)(15) × cos(68) = 2833.4227

∴ f = √2833.4227 = 53.2 feet

* Helio is 53.2 feet from the kite

7 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 :-

$\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}$