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seropon [69]
2 years ago
15

.162q + .035c = 4 Solve for q and c and show work

Mathematics
1 answer:
liberstina [14]2 years ago
7 0

Answer:

q = \frac{5(-7c+800)}{162}

c = \frac{2(-81q + 2000)}{35}

Step-by-step explanation:

First solve for q:

0.162q + 0.035c = 4

0.162q = 4 - 0.035c (move term to other side)

q = \frac{4-0.035c}{0.162} (divide both sides by 0.162 to get it away from q)

q = \frac{5(-7c+800)}{162} (simplify fraction)

Solve for c:

0.162q + 0.035c = 4

0.035c = 4 - 0.162q (move term to other side)

c = \frac{4 - 0.162q}{0.035} (divide both sides by 0.035 to get it away from c)

c = \frac{2(-81q + 2000)}{35} (simplify fraction)

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30 degrees

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If you were to draw a number line, 15 and -15 are both 15 units away from 0. So, 15 + 15 =30

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Which is the graph of f(x) = 2(3)x?
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The last graph.

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This is pretty simple, we know AB is 12 cm and DE is 4 cm. We know ABC has a total of 180 square centimeters. So...

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3 years ago
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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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