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3 years ago

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Fill in missing information to make the equality true: (... +2a)2 = … +12ab2+4a2.

1 answer:

DIA [1.3K]3 years ago

6 0

${ (3b + 2a)}^{2} = 9 {b}^{2} + 12ab + 4 {a}^{2}$

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Graph the exponential function.

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$\bf \: f(x) = \frac{3}{2} \times {2}^{x} \\ \\ = > \bf \: f(0) = \frac{3}{2} \times {2}^{0} = \frac{3}{2} \times 1 = \frac{3}{2} = 1.5 \\ \\ = > \bf \: f(1) = \frac{3}{2} \times {2}^{1} = 3 \\ \\ = > \bf \: f(2) = \frac{3}{2} \times {2}^{2} = 6 \\ \\ = > \bf \: f( - 1) = \frac{3}{2} \times {2}^{ - 1} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} = 0.75 \\ \\ = > \bf \: f( - 2) = \frac{3}{2} \times {2}^{ - 2} = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8} = 0.375$

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3 years ago

The potential solutions to the radical equation are a = −4 and a = −1. Which statement is true about these solutions? The soluti

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Answer:

Both a = −4 and a = −1 are true solutions

Step-by-step explanation:

Given

$\sqrt{a + 5} = a + 3$

$a = -4; a = -1$

Required

The true statement about the solutions

We have:

$\sqrt{a + 5} = a + 3$

Square both sides

$a + 5 = (a + 3)^2$

$a + 5 =a^2 + 6a + 9$

Collect like terms

$a^2 + 6a - a + 9 - 5 = 0$

$a^2 + 5a + 4 = 0$

Expand

$a^2 + 4a + a + 4 = 0$

Factorize

$a(a + 4) + 1(a + 4) = 0$

Factor out a + 4

$(a + 1)(a + 4) = 0$

Split:

$a + 1 = 0; a + 4 = 0$

Solve:

$a =-1; a = -4$

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3 years ago