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mariarad [96]
2 years ago
7

Part B

Mathematics
1 answer:
lidiya [134]2 years ago
4 0

Answer: I don't know

Step-by-step explanation:

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Multiply: 5 • (-3)
slamgirl [31]

Answer:

-15

Step-by-step explanation:

We can factor out a 1 from -3 to get 5\cdot(-3)=5\cdot-1\cdot3=-1(5\cdot3).

We know from simple arithmetic that 5\cdot3=5+5+5=10+5=15.

So, our answer is -1(15)=\boxed{-15}

6 0
3 years ago
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Determine the x- and y-intercepts of the graph of y=1/4x−2 .
Serggg [28]
8 is the x intercept and -2 is the y
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Find the area of the triangle with a= 18.8, b= 11.6, c= 13.8. Round to the nearest tenth.
Anika [276]

Answer:

not rounded is 44.2, rouned to the nearest tenth is 40

Step-by-step explanation:

3 0
3 years ago
What is the sum of 1/2 + 1/4?<br><br> A. 2/8<br> B. 2/6<br> C. 4/8<br> D. 6/8
Pie

<em>3/4 is the total.</em>

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6 0
2 years ago
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
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