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SVEN [57.7K]
3 years ago
11

What should you multiply the top equation by so that y is eliminated when the

Mathematics
2 answers:
VLD [36.1K]3 years ago
8 0
Your answer is D. 2 :)
strojnjashka [21]3 years ago
4 0

Answer:

D

Step-by-step explanation:

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The top left (-5,7)
Middle right (5,1)
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If f(x) = 4x + 2, what is the value of f(3)? <br><br> O 8<br> O 14<br> O 45<br> O 66
lutik1710 [3]

Answer:

14

Step-by-step explanation:

5 0
3 years ago
Como convertir la oración en una ecuación algebraica
zimovet [89]

Ok, entonces creo que sería -6 porque -6 x 3 = -16, no estoy seguro de si esto ayudará o te confundirá más, pero ten un buen día

6 0
3 years ago
Algebra 2, I need help!!! Solve x^2 + 6x + 7 = 0. If you are going to comment in here please know the answer, this is so serious
docker41 [41]

Answer:

Third option

Step-by-step explanation:

We can't factor this so we need to use the quadratic formula which states that when ax² + bx + c = 0, x = (-b ± √(b² - 4ac)) / 2a. However, we notice that b (which is 6) is even, so we can use the special quadratic formula which states that when ax² + bx + c = 0 and b is even, x = (-b' ± √(b'² - ac)) / a where b' = b / 2. In this case, a = 1, b' = 3 and c = 7 so:

x = (-3 ± √(3² - 1 * 7)) / 1 = -3 ± √2

6 0
3 years ago
Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

5 0
3 years ago
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