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Pie
3 years ago
8

What is the correct radical form of this expression?

Mathematics
1 answer:
Sophie [7]3 years ago
4 0

Option A. \sqrt[5]{\left(32 a^{10} b^{\frac{5}{2}}\right)^{2}} is the correct radical form.

Step-by-step explanation:

Step 1:

The numerator of the exponential is kept where it is whereas when converting an exponential into a radical, the denominator becomes the root of the variable.

The numerator of the exponential is 2 while the denominator is 5.

Step 2:

So the numerator is kept where it is while the denominator, 5 becomes the root of the variable.

Out of the given options, options 2, 3 and 4 have square roots so they cannot be the answers. Option 1 has a root to the power of 5, so it is the correct radical form.

So option 1 is the right answer.

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22. If you add a number to a series of data, and it doesn't change the mean, that number MUST equal the mean. Try adding the mean number into it, hopefully it doesn't change the mode or median either.

23. Basically I think (can't quite read all the question!) just work out the circumference of a circle with radius 20 feet (urgh imperial), ie 2*pi*r so 40pi.
4 0
3 years ago
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What is equivalent to 9
frosja888 [35]
What is equivalent to 9?

9/1 = 9

9/1 is equivalent to 9.
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3 years ago
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Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137
Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em>78.70^\circ.

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle \triangle ABC with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles \angle A, \angle B, \angle C

As per Law of cosine:

cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}

Putting the values of a,b and c to find out angles \angle A, \angle B, \angle C.

cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ

cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ

cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ

<em>Camera 2nd has to cover the maximum angle</em>, i.e. 78.70^\circ.

6 0
3 years ago
"let v be the event that a computer contains a virus, and let w be the event that a computer contains a worm. suppose p(v) = 0.1
morpeh [17]

Here we might have to find p(v intersection w) and for that we use the following formula

p(v U w) = p(v)+p(w)-p(v intersection w)

And it is given that p(v) =01.3 , p(w) = 0.04 and p(v U w ) = 0.14 .

Substituting these values in the formula, we will get

0.14 = 0.13 +0.04 -p(v intersection w)

p(v intersection w) =0.13 +0.04 -0.14 = 0.03

So the required answer of the given question is 0.03 .

7 0
3 years ago
Drag expressions to complete each equation.
Maru [420]

Answer:

see explanation

Step-by-step explanation:

Using the rules of exponents

a^{b} × a^{c} = a^{b+c}

\frac{a^{b} }{a^{c} } = a^{b-c}

a^{c} × b^{c} = (ab)^{c}

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3 years ago
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