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mart [117]
2 years ago
15

Gerry is painting a fence. He painted

Mathematics
1 answer:
beks73 [17]2 years ago
4 0
17\20 should be the correct answer
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The equation y = 1/5x represents a proportional relationship. Explain how you can tell the relationship is proportional from the
andre [41]
If a graph is proportional then the line will go through the origin at point (0, 0). If the equation is proportional then it will be in the form of y=kx with no other operations after. The constant of proportionality is another way to say the slope and in your specific equation the slope would be 1/5.
7 0
3 years ago
HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP
romanna [79]

Answer:

D

Step-by-step explanation:

3 0
3 years ago
What is the value of the expression x^2y–y+xy^2–x for x=4 and y=0.25?
stepan [7]

Answer:

0 (zero)

Step-by-step explanation:

{x}^{2} y - y + x {y}^{2}  - x  \\  =  {4}^{2} \times  0.25 - 0.25 + 4 \times  {0.25}^{2}  - 4 \\  = 16 \times  0.25 - 0.25 + 4 \times  0.0625- 4  \\  = 4 - 0.25 + 0.25 - 4 \\  = 0 \\  \\  \purple {\boxed{ \bold{ \therefore \:  {x}^{2} y - y + x {y}^{2}  - x  = 0}}}

8 0
3 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
I need help with 18-27
Gwar [14]

Answer:

18-27= 01

Step-by-step explanation:

18

-27

---------

01

the one cancels out

8 0
3 years ago
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