Q(x)=4x^2-1 find Q(-7)

Answers

Answer 1

Answer:
$q( - 7) = 4x {}^(2) - 1$
so
$4( - 7) {}^(2) - 1$
195

Answer 2

Answer:

$Solution, \mathrm{Isolate}\:x\:\mathrm{for}\:If\left(I,\:x\right)Q\left(x\right)=4x^2-1:\quad x=(√(f\left(I,\:x\right)Q\left(x\right)I+1))/(2),\:x=-(√(f\left(I,\:x\right)Q\left(x\right)I+1))/(2)$

$Steps: If\left(I,\:x\right)Q\left(x\right)=4x^2-1$

$\mathrm{Switch\:sides}, 4x^2-1=If\left(I,\:x\right)Q\left(x\right)$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}, 4x^2-1+1=If\left(I,\:x\right)Q\left(x\right)+1$

$\mathrm{Simplify}, 4x^2=If\left(I,\:x\right)Q\left(x\right)+1$

$\mathrm{Divide\:both\:sides\:by\:}4, (4x^2)/(4)=(If\left(I,\:x\right)Q\left(x\right))/(4)+(1)/(4)$

$\mathrm{Simplify}, x^2=(f\left(I,\:x\right)Q\left(x\right)I+1)/(4)$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=√(f\left(a\right)),\:\:-√(f\left(a\right)), x=\sqrt{(f\left(I,\:x\right)Q\left(x\right)I+1)/(4)},\:x=-\sqrt{(f\left(I,\:x\right)Q\left(x\right)I+1)/(4)},\sqrt{(f\left(I,\:x\right)Q\left(x\right)I+1)/(4)}=(√(f\left(I,\:x\right)Q\left(x\right)I+1))/(2),-\sqrt{(f\left(I,\:x\right)Q\left(x\right)I+1)/(4)}=-(√(f\left(I,\:x\right)Q\left(x\right)I+1))/(2)$

The correct answer is $x=(√(f\left(I,\:x\right)Q\left(x\right)I+1))/(2),\:x=-(√(f\left(I,\:x\right)Q\left(x\right)I+1))/(2)$

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Answers

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What is the product written in scientific notation? (1.4 × 103)(2.5 × 10–6) A. 3.9 × 10–3 B. 3.9 × 10–18 C. 3.5 × 10–3 D. 3.5 × 10–2

Answers

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Hi

1.4 × 10³ × 2.5 × 10⁻⁶ = 1.4 × 2.5 × 10³ × 10⁻⁶ = 3.5 × 10⁻³

1.4 × 2.5 = 3.5

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Answers

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Answers

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Answer:

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