1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
QveST [7]
3 years ago
13

Which is parallel to y = 3x − 5?

Mathematics
1 answer:
nikklg [1K]3 years ago
7 0
The answer is y=3x-6!!!!!
You might be interested in
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
PLEASE ANSWER THIS FAST<br> (x+6)+(5x)=90
Pachacha [2.7K]

Answer:

x=14

Step-by-step explanation:

there is a picture to help you foe the explanation

7 0
3 years ago
Read 2 more answers
Solve the following absolute value equations. Show the solution set and check your answers. |0.3-3/5k|-0.4=1.2
9966 [12]

Answer:

**The equation is not clear, so I have provided both options**

<h3><u>Option 1</u></h3>

\left|0.3-\dfrac{3}{5}k\right|-0.4=1.2

\implies \left|0.3-\dfrac{3}{5}k\right|=1.6

<u>Solution 1</u>

\implies 0.3-\dfrac{3}{5}k=1.6

\implies -\dfrac{3}{5}k=1.3

\implies k=-\dfrac{13}{6}

<u>Solution 2</u>

\implies -(0.3-\dfrac{3}{5}k)=1.6

\implies -0.3+\dfrac{3}{5}k=1.6

\implies \dfrac{3}{5}k=1.9

\implies k=\dfrac{19}{6}

<h3><u>Option 2</u></h3>

\left|0.3-\dfrac{3}{5k}\right|-0.4=1.2

\implies \left|0.3-\dfrac{3}{5k}\right|=1.6

<u>Solution 1</u>

\implies 0.3-\dfrac{3}{5k}=1.6

\implies -\dfrac{3}{5k}=1.3

\implies -3=6.5k

\implies k=-\dfrac{6}{13}

<u>Solution 2</u>

\implies -(0.3-\dfrac{3}{5k})=1.6

\implies -0.3+\dfrac{3}{5k}=1.6

\implies \dfrac{3}{5k}=1.9

\implies 3=9.5k

\implies k=\dfrac{6}{19}

6 0
2 years ago
2.952 + 39.24 = <br>A: 42.192<br>B: 68.76<br>C: 6.876<br>D: 687.6<br>E: 36.288<br>F: 42.76​
kirill [66]

Answer:

The answer to this question is A

4 0
3 years ago
Read 2 more answers
Which description does NOT guarantee that a quadrilateral is a square?
Ivahew [28]
Let's go through the choices one by one

------------------------------------------
Choice A

If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.

In terms of "algebra", you can think
rhombus+rectangle = square

Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.

So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.

------------------------------------------
Choice B

If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.

This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.

------------------------------------------
Choice C

This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.

------------------------------------------
Choice D

As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.

------------------------------------------

Once again, the final answer is choice B


3 0
3 years ago
Other questions:
  • find two consecutive integers such that ten more than twice the smaller is seven less than three times the larger
    15·1 answer
  • A circular cookie cutter has a 2.5 inch radius what is the area of a cookie cut from this cutter
    5·1 answer
  • If Square GHIJ is dilated by a scale factor of five about the center of the square, dilated line K'L' will
    14·2 answers
  • Alexa ran 2/3 of a mile. Caleb ran 5/8 of a mile. Who ran farther?
    13·1 answer
  • A cylinder has a base radius of 9 feet and a
    11·2 answers
  • Which of the following is closest to the diagonal measure of an 18 inch square tile
    5·1 answer
  • HURRY I NEED HELP: Write an equation in point-slope form of the line that passes through the point (3, 5) and has a slope of m=−
    7·1 answer
  • A recipe that serves 8 people requires 4 1/2 cup of flour.
    9·1 answer
  • Simplify the expression:<br> 7(1 + 7x) =
    8·1 answer
  • Mrs Fernandez has a total of 156 red and blue roses.After selling the same number of red and blue roses, she had 1/3 of the numb
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!