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4 years ago

(x-2)(x+2) Find The Product?

2 answers:

5 0

(x-2)*(x+2) = x*(x+2) - 2*(x+2)

(x-2)*(x+2) = x*x + x*2 - 2*x - 2*2

(x-2)*(x+2) = x^2 + 2x - 2x - 4

(x-2)*(x+2) = x^2 - 4

That's the long way to do it. You can use the difference of squares formula as a shortcut

(a-b)*(a+b) = a^2 - b^2

(x-2)(x+2) = x^2 - 2^2

(x-2)(x+2) = x^2 - 4

either way, the final answer is x^2 - 4

jekas [21]4 years ago

4 0

(X-2)(x+2)
So use "f.o.i.l" meaning you multiply each first term, the outside terms, the inside terms and last terms, and then add. You'd get x^2-4 in this example

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3 years ago

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Domain means the values of independent variable(input) which will give defined output to the function.

Given:

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Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

$To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)$

$Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution$

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The height will be positive in the above interval.

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3 years ago

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3 years ago

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