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Julli [10]
3 years ago
7

If the translation of (x,y)->(x-5, y+2) was applied to the triangle ABC below, what would be the coordinates for A' after the

translation?

Mathematics
1 answer:
muminat3 years ago
6 0

Answer:

(-3,0)

Step-by-step explanation:

When translating, anything positive when talking about x would move x to the right. Negative, left, so you would move point a 3 units to the left.

When translating, anything positive when talking about y would move y up. Negative, down, so you would move point a 2 units up.

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The art club has a goal to raise at least $500 by selling paintings to the student body for $15 each. They have already spent $7
natali 33 [55]

Answer:

15x - 70 = 500

Step-by-step explanation:

3 0
3 years ago
James is selecting a marble. James chooses a marble at random and then replaces it. He then selects a second marble at random. W
tresset_1 [31]

Answer:

The probability of selecting a solid black marbles both times;

P = 9/100

Attached is the completed question;

Step-by-step explanation:

Number of solid black marbles = 3

Total number of marbles = 10

The probability of selecting a solid black marble;

P1 = 3/10

With the assumption that the marbles are replaced before next selection.

The probability of selecting a solid black marbles both times;

P = P1 × P1 = 3/10 × 3/10 = 9/100

P = 9/100

7 0
3 years ago
1. Find the midpoint of the segment with endpoints of 6 3i and −12 − 5i. I.
Tju [1.3M]

Answer: what’s is the 5i

Step-by-step explanation:

8 0
3 years ago
Solve the equation.<br> у+ 3 = -у +9<br> y=1<br> y= 3<br> y= 6<br> y= 9
Anna007 [38]
The first y equals -3 and the second y equals -9
8 0
3 years ago
HELP PRECALC DO NOT UNDERSTAND WILL GIVE BRAINLIEST
Olegator [25]

Answer:

  the lower right matrix is the third correct choice

Step-by-step explanation:

Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).

Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.

__

We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.

The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...

  \left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]

Already, we see that the second row matches that in the lower right matrix.

The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.

The correct choices here are the two you have selected, and <em>the lower right matrix</em>.

3 0
4 years ago
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