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Vitek1552 [10]
3 years ago
13

The cube root of 5 times the square root of 5 over the cube root of 5 to the 5th power

Mathematics
1 answer:
Fiesta28 [93]3 years ago
7 0
(cube root of 5)  * sqrt(5)
--------------------------------- = ?
(cube root of 5^5)

This becomes easier if we switch to fractional exponents:


  5^(1/3) * 5^(1/2)         5^(1/3 + 1/2)       5^(5/6)
------------------------ = --------------------- = ------------- = 5^[5/6 - 5/3]
[ 5^5 ]^(1/3)                  5^(5/3)                 5^(5/3) 

Note that 5/6 - 5/3 = 5/6 - 10/6 = -5/6.

                                                            1
Thus,  5^[5/6 - 5/3] = 5^(-5/6)  =  --------------
                                                        5^(5/6)

That's the correct answer.  But if you want to remove the fractional exponent from the denominator, do this:

    1            5^(1/6)         5^(1/6)
---------- * ------------- = --------------    (ANSWER)
5^(5/6)       5^(1/6)              5
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Assume V and W are​ finite-dimensional vector spaces and T is a linear transformation from V to​ W, T: Upper V right arrow Upper
scZoUnD [109]

Answer:

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

Step-by-step explanation:

Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that  

T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)

so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.  

Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that  

c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0

T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0

T(c_1v_1+ c_2v_2 ... c_pv_p) = 0  

But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

8 0
3 years ago
Math problem I need help giving brainly! Super easy
Alborosie

Answer:

answer is -3

Step-by-step explanation:

one solution

6 0
2 years ago
Answer 15 points <br> ( don’t answer if you don’t the answer )
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Answer:

V=1/3hπr². Basically, 1/3 x height x 3.14 (pi) x radius squared.

Step-by-step explanation:

The formula of figure 2 would be 1/3 x 15 x 3.14 x 8 squared.  The volume of figure 2 would be 1005.30965

the formula for figure 3 would be 1/3 x 15 x 3.14 x 4 squared. The volume of it would be 251.32741

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8 0
2 years ago
I really need help on this
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Do you know the subject that the problem is about ?
6 0
3 years ago
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What type of angle is this?​
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Answer:

Obtuse Angle

Step-by-step explanation:

It is greater than 90˚

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