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amid [387]
3 years ago
14

Match the function with the graph

Mathematics
2 answers:
Gre4nikov [31]3 years ago
5 0

Answer:

Choice A is the correct answer

Step-by-step explanation:

First things first, the function has a y-axis symmetry and is thus an even function. Choice B and D are automatically incorrect since the sine function is always odd. I employed Desmos graphing tool to plot the two remaining functions, and choice A matched the given graph

icang [17]3 years ago
3 0

Answer:

Option: a is correct

a) y= -cos x

Step-by-step explanation:

We could clearly see that the graph of the function passes through a point on the y-axis which is :

(0,-1)

i.e. when x=0 the value of the function y= -1

So we will check in each of the following options by putting x=0 and check whether y= -1 or not.

b)

y=sin x

on putting x=0 we get:

y=sin 0=0≠ -1.

Hence, option b is incorrect.

c)

y= cos x.

we put x=0

y= cos 0=1≠ -1.

Hence, option c) is incorrect.

d)

y= -sin x

we put x=0 to obtain:

y= -sin 0=0≠ -1.

Hence, option d is incorrect.

Hence we are left with option a i.e.

The graph represents the function:

y= -cos x.  (Hence, option a is correct)

( Also it satisfies at x=0

since, y= -cos 0= -1 )

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<em>solution</em>

<em>or</em><em>,</em><em> </em><em> </em><em> </em><em> </em><em>2 x + y </em><em>=</em><em> 5</em><em> y - </em><em>8</em><em> </em><em> </em><em> </em><em> </em><em> </em><em>being opposite side of </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>parallelogr</em><em>a</em><em>m</em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em>or </em><em>,</em><em> </em><em>2x </em><em>=</em><em>5 y </em><em>-</em><em> </em><em>y</em><em> </em><em>-</em><em> </em><em>8</em>

<em>o</em><em>r</em><em>,</em><em> </em><em> </em><em>2x </em><em>=</em><em> </em><em>4 y -</em><em> 8</em>

<em>or</em><em>,</em><em> </em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>4</em><em>y</em><em> </em><em>-</em><em> </em><em>8</em><em>/</em><em>2</em>

<em>or</em><em>,</em><em>. </em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>2</em><em>y</em><em>-</em><em> </em><em>4</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>therefore</em><em>,</em><em> 2x + y</em><em> is equal to</em><em> </em><em>5 y - 8</em><em> </em><em>Being</em><em> the </em><em>opposite side of a parallelogram</em>

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<em>or</em><em>,</em><em> </em><em>3y</em><em> </em><em>+</em><em> </em><em>2</em><em> </em><em>x-4x</em><em> </em><em>=</em><em> </em><em>-</em><em>3</em>

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