How do you evaluate $7 + (- 4 x^{2})$ $7+(-4x^2)$ for $x = 0$ $x=0$ ?

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Answer 1 · 2017-06-19T14:27:58

7

$7 + (- 4 x^{2})$ $7+(-4x^2)$ for $x = 0$ $x=0$

Solution
Since $x = 0$ $x=0$ you have to substitute $0$ $0$ to anywhere you see $x$ $x$

$:.$ $7+(-4(0^2))$

$7+(-4(0))$

$7+(-0)$

$7-0$

$Answer$ $->$ $7$

Answer 2 · 2017-06-19T14:31:54

The answer is 7.

All you have to do is simply substitute 0 in for x:
$7+(-4(0^2))$
$7+(-4(0))$
$7+0=7$

How do you evaluate 7+(-4x^2)7+(−4x2)7+(-4x^2) for x=0x=0x=0?