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Goryan [66]
3 years ago
13

Find all the real square roots of 0.0064.

Mathematics
2 answers:
Anika [276]3 years ago
7 0

Answer: 1)  Third option is correct.

2)  Second option is correct.

Step-by-step explanation:

1) Since we have given that

\sqrt{0.0064}

We need to find all the real square roots.

First we will write it in the following way:

\sqrt{0.0064}\\\\=\sqrt{\frac{64}{10000}}\\\\=\frac{\sqrt{64}}\sqrt{10000}}\\\\=\pm\frac{8}{100}\\\\=\pm 0.08\\\\=0.08,-0.08

Hence, the real square roots of 0.0064 are 0.08 and -0.08.

Hence, Third option is correct.

2) The real number root will be of 1.1 as it is the only positive number among the options for the real number root.

\sqrt{1.1}=1.048

So, Second option is correct.

Degger [83]3 years ago
3 0
<span>(a ^ b) ^ c = a ^ (bc) using this formula
</span><span> 64 * 10^-4.
</span>sqrt(64 * 10^-4) 
<span>sqrt(64) * sqrt(10^-4) </span>
<span>=+/- 8 * (10^-4)^1/2 </span>
<span>=+/-8 * (10)^(-4 * 1/2) </span>
<span>=+/-8 * 10^(-2) </span>
<span>=+/-8 * 0.01 </span>
<span>=+/- 0.08 
</span>hope it helps
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a rider sits on a motorcycle . the motorcyle has amass of 237 kilograms . the rider has a mass of 89 kilograms. what is the tota
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4 0
3 years ago
the distant between two towns is kilometers there are approximately 8 kilometers in 5 miles which measurement is closest to the
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3 years ago
Determine the number of possible solutions for a triangle with A= 30 a=20 and b=16
Bess [88]
The answer is: Two possible solutions which are (0.53, 37.19)

Explanation:

Given:
A = 30° 
<span>a = 20 </span>
<span>b = 16 </span>

Now use the law of Cosines:
a² = b² + c² - 2bc*cos(A)

Plug in the values:

20² = 16² + c² - (2*(16)*c*cos(30))
<span>400 = 256 + c² - 32c(0.866) </span>
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<span>c² - 27.71c = 400 - 256 </span>
<span>c² - 27.71c = 144 </span>
<span>c² - 27.71c + 191.96 = 144 + 191.96 </span>
<span>(c - 18.86)² = 335.96 </span>
<span>c - 18.86 = √336.95 </span>
<span>c - 18.86 = ± 18.33 </span>
<span>c = 18.86 ± 18.33 </span>

<span>If c = 18.86 + 18.33, then </span><span>c = 37.19 </span>

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4 0
3 years ago
Read 2 more answers
I just need the extraneous root
gregori [183]

Answer:

Step-by-step explanation:

2+\sqrt{2x-3} =\sqrt{x+7}  ~~~(1)\\ x \ge -7\\and ~x\ge \frac{3}{2} \\squaring ~(1)\\4+2x-3+4\sqrt{2x-3} =x+7\\4\sqrt{2x-3} =x+7-2x-4+3\\4\sqrt{2x-1} =-x+6\\squaring\\16(2x-3)=x^2-12x+36\\32x-48=x^2-12x+36\\x^2-12x-32x+36+48=0\\x^2-44x+84=0\\\\x^2-2x-42x+84=0\\x(x-2)-42(x-2)=0\\(x-2)(x-42)=0\\x=2,42\\x=2 is a solution.\\whereas ~x=42 ~is~an~extraneous ~root.\\as~2+\sqrt{2*42-3} =\sqrt{42+7} \\2+\sqrt{81} =\sqrt{49} \\2+9=7\\or 11=7\\which~is~not~true.

8 0
3 years ago
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