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Pachacha [2.7K]
3 years ago
11

What rule describes the translation? Write tule as an ordered pair

Mathematics
1 answer:
RSB [31]3 years ago
8 0
For A (x+6), (y-2)
For B (x+6), (y-2)
For C (x+2), (y+2)
Good luck
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Suppose T: ℝ3→ℝ2 is a linear transformation. Let U and V be the vectors given below, and suppose that T(U) and T(V) are as given
ira [324]

Answer:

T(3u+2v)=(-10,29)

Step-by-step explanation:

T is a linear transformation, hence it is homogeneous (T(cr)=cT(r) for all real c and r∈ℝ³) and additive (T(r+s)=T(r)+T(s), for all r,s∈ℝ³). Apply these properties with r=3u and s=2v to obtain:

T(3u+2v)=T(3u)+T(2v)=3T(u)+2T(v)=3(-2,5)+2(-2,7)=(-6,15)+(-4,14)=(-10,29)

We don't have an explicit definition of T, so it's more difficult to compute T(3u+2v) directly without using these properties.

8 0
3 years ago
(a-12)x = a+8 in the equation to the left, a is a constant. if the equation has no solutions, what is the value of a?
ohaa [14]

Answer:

<h2>The value of a is 12.</h2>

Step-by-step explanation:

The equation is given by (a - 12)x = a + 8.

In order to get the solution of the equation, that is the value of x, the above equation is needs to be divided by a - 12. (a - 12) will be in the denominator.

The solution can not be found, if the value of a - 12 will be 0, that is a = 12.

6 0
3 years ago
Select all the ordered pairs that are solutions to the equation 5x + 2y = -8
Archy [21]

Answer: (1,-2) (0,-4) (-4,0)

Step-by-step explanation:

3 0
3 years ago
Ben bowled 139 and
QveST [7]

Answer:

187

Step-by-step explanation:

Were gonna start by writing an equation (139+184+x)/3=170 because were adding a 3rd value to the average and x is going to be our value that were looking for. If you solve that, you get 187. Hope this helps.

3 0
3 years ago
Read 2 more answers
How to solve this??<br><br> (r^-7b^-8) ^0<br> -------------<br> (t^-4w)
professor190 [17]

keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.


\bf ~~~~~~~~~~~~\textit{negative exponents}&#10;\\\\&#10;a^{-n} \implies \cfrac{1}{a^n}&#10;\qquad \qquad&#10;\cfrac{1}{a^n}\implies a^{-n}&#10;\qquad \qquad a^n\implies \cfrac{1}{a^{-n}}&#10;\\\\[-0.35em]&#10;\rule{34em}{0.25pt}\\\\&#10;\cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}

6 0
3 years ago
Read 2 more answers
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