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Trava [24]
3 years ago
13

Which expression is equivalent to ^3√x^5y?a)x5/3b)x5/3y1/3c)x3/5 yd)x3/5 y3​

Mathematics
1 answer:
Illusion [34]3 years ago
7 0

Answer:

(x^5y)^{\frac{1}{3}}= x^{\frac{5}{3}} \times y^{\frac{1}{3}}

Step-by-step explanation:

We are given our expression as

\sqrt[3]{x^5y}

The property of exponent says that

\sqrt[n]{x} =x^{\frac{1}{n}}

Hence

\sqrt[3]{x^5y}= (x^5y)^{\frac{1}{3}}

Another property says

(ab)^n=a^n \times b^n

Hence

(x^5y)^{\frac{1}{3}}= x^{\frac{5}{3}} \times y^{\frac{1}{3}}

Hence option B is correct.

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Identify the axis of symmetry of the given quadratic<br> y= -3x^2 - 12
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Answer:

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0 = x

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Step-by-step explanation:

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2 years ago
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8 0
3 years ago
Find equations of the spheres with center(1, −1, 6)that touch the following planes.a) xy-planeb) yz-planec) xz-plane
Misha Larkins [42]

Given :

Center of sphere , C( 1 , -1 , 6 ) .

To Find :

Find equations of the spheres with center (1, −1, 6) that touch the following planes.a) xy-plane b) yz-plane c) xz-plane .

Solution :

a)

Distance of the point from xy-plane is :

d = 6 units .

So , equation of circle with center C and radius 6 units is :

(x-1)^2+(y-(-1))^2+(z-6)^2=6^2\\\\(x-1)^2+(y+1)^2+(z-6)^2=36

b)

Distance of point from yz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1

c)

Distance of point from xz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1

Hence , this is the required solution .

4 0
2 years ago
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