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natulia [17]
3 years ago
9

How do I solve this equation?

Mathematics
1 answer:
shusha [124]3 years ago
3 0

Answer:

x≤-4 or x≥0.5

Step-by-step explanation:

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Change the expression to a single square root, or its opposite:
aleksley [76]

Answer:

a)2\sqrt{2}=\sqrt{8}

b)-7\sqrt{3} =-\sqrt{147}

c)\frac{1}{3} \sqrt{18b}  =\sqrt{2.b}

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e)-6\sqrt{2a}  =-\sqrt{72a}

f)-0.1\sqrt{200c}=-  \sqrt{2c

Step-by-step explanation:

a) 2\sqrt{2}=( \sqrt{2}  )^2.\sqrt{2}=(\sqrt{2})^3 = \sqrt{2^3} =\sqrt{8}

b)-7\sqrt{3} =-(\sqrt{7} )^2\sqrt{3} =-\sqrt{7^2.3} =-\sqrt{147}

c)\frac{1}{3} \sqrt{18b} =\frac{1}{3} \sqrt{9.2.b} =\frac{1}{3} \sqrt{3^2.2.b} =\frac{1}{3} \times 3\sqrt{2.b} =\sqrt{2.b}

d)5\sqrt{y} =\sqrt{5^2} \sqrt{y}=\sqrt{25y}

e)-6\sqrt{2a} =-\sqrt{6^2}\sqrt{2a}  = -\sqrt{36.2a} =-\sqrt{72a}

f)-0.1\sqrt{200c}=-\frac{1}{10}  \sqrt{10^2.2c} =-\frac{1}{10}\times10  \sqrt{2c}=-  \sqrt{2c

4 0
3 years ago
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