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Anton [14]
3 years ago
9

Find the perimeter and area of a triangle with the vertices and (3,0) and (6,4)

Mathematics
1 answer:
madreJ [45]3 years ago
4 0
You have to use the distance formula which is d=√(x₂-x₁)²+(y₂-y₁) to give you the distance between the points then solve for perimeter and area once you have the distance

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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
PLS HURRY!
Ratling [72]
A is the correct answer 3 terms and a degree of 9:)
4 0
2 years ago
Carole purchased 7/8 of a pound of chocolate chips to make cookies. On the way home from school she got hungry and ate 1/4 of th
AURORKA [14]
Answer is below..................

4 0
3 years ago
How can you use 4s fact to find 7x8? Give the product in your explanation.
Mkey [24]
So 4's facts

we know that 4 times 2=8 and the associative property of multiplication says
a(bc)=(ab)c

so 7(8) (means 7 times 8) means 7((4)(2)) and associative say we can group them and say (7)(4)(2) so (4)((7)(2)) or 4(14) and 4 times 14=56

7 times 8=56
7 0
3 years ago
Determine where to place the decimal point in the dividend and divisor so that the quotient is between 23 and 25
timama [110]

Answer:

Place the decimal before the last digit in both number i.e. 1602.3 and 65.4

Step-by-step explanation:

We are given the numerical expression 16023÷654.

Here, we have,

Dividend = 16023

Divisor = 654

it is required that the quotient of the division to be between 23 and 25.

If we take the numbers,

1602.3 and 65.4

This gives us,

\frac{1602.3}{65.4}=24.5.

Hence, placing the decimal before the last digit in both number will give the desired result.

7 0
2 years ago
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