All categories

2 years ago

Look at the graph below. Is it a function or not a function? Explain your reasoning.

Mathematics

1 answer:

navik [9.2K]2 years ago

6 0

Answer:

not a function because it dont cross the axis

You might be interested in

Valeria's office recycled a total of 44 kilograms of paper over 4 weeks. How many weeks will it take Valeria's office to recycle

diamong [38]

The answer would be 6 weeks.

First, you do 44 divided by 4 to get 11 kg per week. Now you divide 66 by 11 to get 6 weeks.

8 0

2 years ago

If Amanda walks at an average speed of 2.72 miles per hour how long will it take her to walk 6.8 miles

defon

Two and a half hours

8 0

3 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

Algebra 1 Question Brainliest given

agasfer [191]

D hope this helps :) good luck

4 0

3 years ago

Read 2 more answers

1)a chord of length 18cm midway the radius of a circle. calculate the radius of the circle correct to 1d.p. 2)if two parallel ch

kykrilka [37]

Part 1:

Given that the length of the chord is 18 cm and the chord is midway the radius of the circle.

Thus, half the angle formed by the chord at the centre of the circle is given by:

$\cos\theta=\frac{\left( \frac{1}{2} r\right)}{r}= \frac{1}{2} \\ \\ \Rightarrow\theta=\cos^{-1}\left( \frac{1}{2} \right)=60^o$

Now,

$\sin60^o= \frac{9}{r} \\ \\ \Rightarrow r= \frac{9}{\sin60^o} =10.392$

Therefore, the radius of the circle is 10.4 cm to 1 d.p.

Part 2I:

Given that the radius of the circle is 10 cm and the length of chord AB is 8 cm. Thus, half the length of the chord is 4cm. Let the distance of the mid-point O to /AB/ be x and half the angle formed by the chord at the centre of the circle be θ, then

$\sin\theta= \frac{4}{10} = \frac{2}{5} \\ \\ \theta=\sin^{-1}\left( \frac{2}{5} \right)=23.6^o$

Now,

$\cos23.6^o= \frac{x}{10} \\ \\ \Rightarrow x=10\cos23.6^o=9.165\approx9.2cm$

Part 2II:

Given that the radius of the circle is 10cm and the angle distended is 80 degrees. Let half the length of chord CD be y, then:

$\sin40^o= \frac{y}{10} \\ \\ \\ \Rightarrow y=10\sin40^o=6.428$

Thus, the length of chord CD = 2(6.428) = 12.856 which is approximately 12.9 cm.

3 0

3 years ago

Look at the graph below. Is it a function or not a function? Explain your reasoning.​

Look at the graph below. Is it a function or not a function? Explain your reasoning.