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Fittoniya [83]
4 years ago
10

1. Demonstrate the validity of the following identities by means of truth tables: (5 pts each) a. DeMorgan’s theorm for three va

riables: (x+y+z)’ = x’y’z’ And (xyz)’ = x’+y’+z’ b. The distributive law: x+yz = (x+y)(x+z)

Mathematics
1 answer:
ollegr [7]4 years ago
3 0

Answer:

All the three statements presented are indeed valid and satisfy the De Morgan's theorem.

Step-by-step explanation:

The truth tables are presented in the attached images.

a) The statement (X + Y + Z)' = (X'Y'Z') has its proof in the first image attached.

b) The statement (XYZ)' = (X' + Y' + Z') has its proof in the second image attached.

c) The statement (X + YZ) = (X + Y)(X + Z) has its proof in the third image attached.

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Which of these strategies would eliminate a variable in the system of equations?
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Step-by-step explanation:

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Futhe Mathematics<br><img src="https://tex.z-dn.net/?f=%28Cos%20%7B%7D%5E%7B4%7Dt%20-Sin%20%7B%7D%5E%7B4%7Dt%20%29%20%20%5Cdiv%2
Nastasia [14]

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }

We can write the numerator as ,

\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }

Recall the identity ,

\longrightarrow (a-b)(a+b)=a^2-b^2

Using this we have ,

\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}

Again , as we know that ,

\longrightarrow sin^2\phi + cos^2\phi = 1

Therefore we can rewrite it as ,

\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}

Again using the first identity mentioned above ,

\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}

Or else we can also write it using ,

\longrightarrow cos2\phi = cos^2\phi - sin^2\phi

Therefore ,

\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}

And we are done !

\rule{200}{4}

Additional info :-

<em>D</em><em>e</em><em>r</em><em>i</em><em>v</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>o</em><em>f</em><em> </em><em>c</em><em>o</em><em>s</em><em>²</em><em>x</em><em> </em><em>-</em><em> </em><em>s</em><em>i</em><em>n</em><em>²</em><em>x</em><em> </em><em>=</em><em> </em><em>c</em><em>o</em><em>s</em><em>2</em><em>x</em><em> </em><em>:</em><em>-</em>

We can rewrite cos 2x as ,

\longrightarrow cos(x + x )

As we know that ,

\longrightarrow cos(y + z )= cosy.cosz -  siny.sinz

So that ,

\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x)

On simplifying,

\longrightarrow cos(x+x) = cos^2x - sin^2x

Hence,

\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}

\rule{200}{4}

7 0
2 years ago
I need help with this math
prisoha [69]

Answer:

EF =11 ft

Step-by-step explanation:

if CB = 11ft then EF will also be 11 ft since both triangles are congruent to each other

6 0
2 years ago
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