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
Zanzabum
2 years ago
11

Peter is plotting equivalent ratios on the coordinate plane. He has plotted these points. Equivalent Ratios If he plots a point

with an x-value of 9, what will be the y-value? Use the interactive graph to find the equivalent ratio. y =
Mathematics
2 answers:
Nuetrik [128]2 years ago
7 0

Answer:3

Step-by-step explanation:

Trust

german2 years ago
3 0

Answer:

the answer is 3

Step-by-step explanation:

it just is

You might be interested in
Question 9 (5 points)
DochEvi [55]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
Pls help I would really appreciate it
Nadusha1986 [10]

Answer:

the ANWER IS B

Step-by-step explanation:

It B BECAUSE I AM SMART

8 0
2 years ago
Read 2 more answers
Answer the photo below thanks
nataly862011 [7]

Answer:

octagon, because 4 + 4 = eight sides, and octagons have 8 sides

Step-by-step explanation:

Hope that helps!

7 0
3 years ago
Read 2 more answers
Select the correct answer. which expression is equivalent to the given expression? assume the denominator does not equal zero. a
Alexxandr [17]

If the denominator does not equal 0, the equivalent expression of \frac{14x^4y^6}{7x^8y^2} is \frac{2y^{4}}{x^{4}}\frac{x^{10} y^{14}}{729}

<h3>How to determine the equivalent expression?</h3>

The expression is given as:

\frac{14x^4y^6}{7x^8y^2}

Divide 14 by 7

\frac{14x^4y^6}{7x^8y^2} = \frac{2x^4y^6}{x^8y^2}

Apply the law of indices

\frac{14x^4y^6}{7x^8y^2} = \frac{2y^{6-2}}{x^{8-4}}

Evaluate the differences in the exponents

\frac{14x^4y^6}{7x^8y^2} = \frac{2y^{4}}{x^{4}}

Hence, the equivalent expression of \frac{14x^4y^6}{7x^8y^2} is \frac{2y^{4}}{x^{4}}

Read more about equivalent expressions at:

brainly.com/question/2972832

#SPJ4

4 0
2 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
Other questions:
  • brainly Chase has been playing a game where he can create towns and help his empire expand. Each town he has allows him to creat
    13·1 answer
  • Please help ;-;, I really need it-
    9·1 answer
  • Solve the system of inequalities by graphing find all corner points
    7·2 answers
  • An engineer is building a suspended platform to raise bags of cement. The platform has a mass of 200 kg, and each bag of cement
    14·1 answer
  • What is the solution to 4x-8=12 please explain.
    11·2 answers
  • Will mark braniest 2 left also add me i help
    14·2 answers
  • What is the measure of angle z in this figure?
    6·2 answers
  • i need help on this it’s asking which of the following is the equation of the linear function (f)x shown graphed below
    15·1 answer
  • When you multiply 2 decimals, how do you know where to place the comma in the product?
    6·2 answers
  • HELPPPPP ASAP!!!! I’m on a limit
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!