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Gnoma [55]
3 years ago
13

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} } }

<u>Solving the given expression</u><u> </u><u>:</u><u>-</u><u> </u>

\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)}

<u>Using</u><u> </u><u>(</u><u>a</u><u> </u><u>+</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>(</u><u>a</u><u> </u><u>-</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>a²</u><u> </u><u>-</u><u> </u><u>b²</u>

\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
Y = 1/2x + 4
Mademuasel [1]

no solution.

The two lines are parallel

(same slope but different y intercepts)

6 0
2 years ago
Question 6(Multiple Choice Worth 5 points)
9966 [12]
Question 6: $38.82
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3 0
3 years ago
Read 2 more answers
What is the area of a bedroom thats 12 ft in length and 10 ft in width?
ludmilkaskok [199]

Answer:

120 sq ft.

Step-by-step explanation:

REMEMBER: Area is found by multiplying one side, by the other side that isn't the same length as it.

Multiply 12 by 10 to get 120

120 is the area of the bedroom

3 0
2 years ago
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