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anastassius [24]
3 years ago
7

NEED HELP WITH THIS PLSSS!!!!

Mathematics
1 answer:
rewona [7]3 years ago
5 0

Answer:

use the formula with the width and height number

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Show that the transformation T defined by<br>T(x1, x2) = (2x1 - 3x2, X1+4, 5x2) is not linear.<br>​
erastovalidia [21]

Let <em>u</em> = (1, 0) and <em>v</em> = (0, 1). Then

<em>T</em> (<em>u</em>) = (2*1 - 3*0, 1 + 4, 5*0) = (2, 5, 0)

<em>T</em> (<em>v</em>) = (2*0 - 3*1, 0 + 4, 5*1) = (-3, 4, 5)

=>   <em>T</em> (<em>u</em>) + <em>T</em> (<em>v</em>) = (-1, 9, 5)

but

<em>T</em> (<em>u</em> + <em>v</em>) = <em>T</em> (1, 1) = (2*1 - 3*1, 1 + 4, 5*1) = (-1, 5, 5)

=>   <em>T</em> (<em>u</em> + <em>v</em>) ≠ <em>T</em> (<em>u</em>) + <em>T</em> (<em>v</em>)

which means <em>T</em> does not preserve addition, so it is not linear.

4 0
3 years ago
The system has......
Flura [38]

Answer:

No solution

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Suppose that 70% of all Americans agree that a candidate is not trustworthy. A survey finds the 116 out of 200 agree that a cand
MArishka [77]

Answer:

Prob = 0.00009 = 0.009%  , Less Likely

Step-by-step explanation:

Using Binomial Distribution, Prob = N c r . ( P )^ r . ( Q ) ^ (n -r)   .

,P = success probability, ie candidate considered be trustworthy = 70% = 0.7; N = trials = 200 ; Q = failure probability, ie non trustworthy = 1 - 0.7 = 0.3  r = no. of success = 116

Pr = 200c116 . (0.7)^116 . (0.3)^(200 - 116) = 200c116 (0.7)^116 (0.3)^84 = 0.00009 = 0.009%

Less probability means Less Likely

5 0
3 years ago
Please answer this question
Gelneren [198K]
A=2(LW+LH+WH)

A=2((7/8)(1/3)+(7/8)(2/5)+(1/3)(2/5))

A=2(7/24+14/40+2/15)

A=14/24+28/40+4/15

A=7/14+7/10+4/15  210

A=(105+147+56)/210

A=308/210

A=(210+98)/210

A=1 98/210

A=1 7/15
8 0
3 years ago
Read 2 more answers
Please anyone help
serg [7]
5.46 to 3 sig figs means the range is {5.455,5.465} and 17.74 means the range {17.735,17.745}.
p=q²/r has a maximum value when q=5.465 and r=17.735 and a minimum value when 5.455 and r=17.745.
So the range of p is 1.6769 to 1.6840. When we have 2 decimal places we get p=1.68 which accommodates the maximum and minimum values of the range. So 2 decimal places is a suitable degree of accuracy, or we could say 3 significant figures.
7 0
3 years ago
Read 2 more answers
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