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2 years ago

Solving systems of equations by substitution y=4x and 6x-4y=-10

Mathematics

1 answer:

slava [35]2 years ago

5 0

The answer I got was (1,4)

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Without using a calculator, determine the number of real zeros of the function

kumpel [21]

The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

<h3>How to determine the number of real zeros?</h3>

The equation of the function is given as:

$f(x) = x^3 + 4x^2 + x - 6$

Expand the function

$f(x) = x^3 + 5x^2 - x^2 + 6x - 5x - 6$

Reorder the terms

$f(x) = x^3 + 5x^2 + 6x - x^2 - 5x - 6$

Factor the expression

$f(x) = x(x^2 + 5x + 6) -1(x^2 + 5x + 6)$

Factor out x -1

$f(x) = (x^2 + 5x + 6)(x -1)$

Expand

$f(x) = (x^2 + 3x + 2x + 6)(x -1)$

Factorize

$f(x) = [x(x + 3) + 2(x + 3)](x - 1)$

Factor out x + 2

$f(x) = (x + 3)(x + 2)(x- 1)$

The function has been completely factored and it has 3 linear factors

Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

Read more about functions at:

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2 years ago

Can someone please help me with these questions!! Thanks,

fiasKO [112]

6a. 1 - 2sin(x)² - 2cos(x)² = 1 - 2(sin(x)² +cos(x)²) = 1 - 2·1 = -1

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6e. 3sin(x) + tan(x)cos(x) = 3sin(x) + (sin(x)/cos(x))cos(x) = 3sin(x) +sin(x) = 4sin(x)

6g. 1 - cos(x)²tan(x)² = 1 - cos(x)²·(sin(x)²)/cos(x)²) = 1 -sin(x)² = cos(x)²

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