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lakkis [162]
2 years ago
10

A square of an even number is always even.True or false?​

Mathematics
2 answers:
kondaur [170]2 years ago
8 0

Answer:

True

Step-by-step explanation:

The square of an even number is always even.

Examples:

8² = 64

64² = 4096

6² = 36

13040² = 170,041,600

aksik [14]2 years ago
5 0

Answer:

Mostly likely it's True

Step-by-step explanation:

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statuscvo [17]

Answer:

There are no solutions to the inequality.

Step-by-step explanation:

|x - 3| < x – 3

1. Separate the inequality into two separate ones.

(1) x – 3 < x – 3

(2) x – 3 < -(x – 3)  

2. Solve each equation separately

(a) Equation (1)

\begin{array}{rcl}x - 3 & < & x - 3\\x & < & x\\\end{array}\\\text{This is impossible. No solutions exist.}

(b) Equation (2)

\begin{array}{rcl}x - 3 & < & -(x - 3)\\x - 3 & < & -x + 3\\x &

For example, if x = 0, we get  

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4 0
3 years ago
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Virty [35]
To factor, you split the equation so the factors multiply to get the whole equation.
(x-2)(x+2)

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3 years ago
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How many complex zeros does the polynomial function have? <br><br> F(x)=2x^4 +5x^3 - x^2 +6x-1
klio [65]

Answer:

Two complex roots.

Step-by-step explanation:

F(x)=2x^4 +5x^3 - x^2 +6x-1

is a polynomial in x of degree 4.

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Use remainder theorem to find the roots of the polynomial.

F(0) = -1 and F(1) = 2+5-1+6-1 = 11>0

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Thus there is a real root between 0 and 1.

Similarly by trial and error let us find other real root.

F(-3) = -1 and F(-4) = 94

SInce there is a change of sign, from -4 to -3 there exists a real root between -3 and -4.

Other two roots are complex roots since no other place F changes its sign

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