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1 year ago

Which graph represents a function?

Mathematics

2 answers:

Agata [3.3K]1 year ago

8 0

The fourth one because to know if it is a function or not it has to pass the vertical line test the first one when you draw a vertical line some of the lines repeats so does the second and third but the last one when you draw a vertical line it doesn't repeat s that means its a function

Kay [80]1 year ago

3 0

Answer: The correct graph is the last one. Its image is attached.

Step-by-step explanation:

We are given the figures of four different graphs. We are to select the one which represents a function.

On a graph paper, a function is defined as follows:

<u>FUNCTION:</u> A graph on the co-ordinate plane represents a function if one value of 'x' corresponds to only one value of 'y'

That is, one value of 'x' cannot give two different values of 'y'.

We can see that

In graph (1), x = 1 corresponds to y = 3, -2.

In graph (2), x = 2 corresponds to y = 2, -2.

In graph (3), x = -3 corresponds to y = 2, -2.

So, these three graphs does not represent functions.

But, in graph (4), one value of 'x' gives only one value of 'y'.

For example, if x = 1, then y = 0

if x = 0, then y = 0, etc.

Therefore, graph (4) represents a function. Its image is attached.

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A soccer ball kicked by a Real Madrid midfielder leaves the player's foot at a speed of 48 mph at an angle θ above the horizonta

kodGreya [7K]

<h2>For maximum range, θ = π/4 </h2>

Step-by-step explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = -u sin θ

Acceleration = -g

Substituting

v = u + at

-u sin θ = u sin θ - g t

$t=\frac{2usin\theta }{g}$

This is the time of flight.

Consider the vertical motion of ball till maximum height,

We have equation of motion v² = u² + 2as

Initial velocity, u = u sin θ

Acceleration, a = -g

Final velocity, v = 0 m/s

Substituting

v² = u² + 2as

0² = u²sin² θ + 2 x -g x H

$H=\frac{u^2sin^2\theta }{2g}$

This is the maximum height reached,

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{2usin\theta }{g}+0.5\times 0\times (\frac{2usin\theta }{g})^2\\\\s=\frac{2u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{g}$

This is the range.

For maximum range

sin 2θ = 1

2θ = π/2

θ = π/4

For maximum range, θ = π/4

8 0

1 year ago

In an examination, a candidate has to score 40% marks to pass. Anil scores 22 and failed by 8 marks. What is the full marks?

alexgriva [62]

Answer:

Step-by-step explanation:

Let full marks be x

Then pass marks is 40% of x which is 0.4x

As per given, we have

22= 0.4x - 8

Solving for x

0.4x = 30

x = 30/0.4

x = 75

So full marks is 75

8 0

2 years ago

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

###

Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

3 0

1 year ago

A cylindrical roller 2.5 m in length, 1.5 m in radius when rolled on a road was found to cover the area of 16500 m2 . How many r

eduard

Answer:

701 revolutions

Step-by-step explanation:

Given: Length= 2.5 m

Radius= 1.5 m

Area covered by roller= 16500 m²

Now, finding the Lateral surface area of cylinder to know area covered by roller in one revolution of cylindrical roller.

Remember; Lateral surface area of an object is the measurement of the area of all sides excluding area of base and its top.

Formula; Lateral surface area of cylinder= $2\pi rh$

Considering, π= 3.14

⇒ lateral surface area of cylinder= $2\times 3.14\times 1.5\times 2.5$

⇒ lateral surface area of cylinder= $23.55 \ m^{2}$

∴ Area covered by cylindrical roller in one revolution is 23.55 m²

Next finding total number of revolution to cover 16500 m² area.

Total number of revolution= $\frac{16500}{23.55} = 700.6369 \approx 701$

Hence, Cyindrical roller make 701 revolution to cover 16500 m² area.

8 0

1 year ago

Let p denote the proportion of students at a particular university that use the fitness center on campus on a regular basis. For

9966 [12]

Answer:

a) P-value = 0.0968

b) P-value = 0.2207

c) P-value = 0.0239

d) P-value = 0.0040

e) P-value = 0.5636

Step-by-step explanation:

As the hypothesis are defined with a ">" sign, instead of an "≠", the test is right-tailed.

For this type of test, the P-value is defined as:

$P-value=P(z>z^*)$

being z* the value for each test statistic.

The probability P is calculated from the standard normal distribution.

Then, we can calculate for each case:

(a) 1.30

$P-value=P(z>1.30) = 0.0968$

(b) 0.77

$P-value=P(z>0.77) = 0.2207$

$P-value=P(z>1.98) = 0.0239$

(d) 2.65

$P-value=P(z>2.65) = 0.0040$

(e) −0.16

$P-value=P(z>-0.16) = 0.5636$

7 0

1 year ago