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2 years ago

For which independent value do the equations generate the same dependent value?

Mathematics

1 answer:

nalin [4]2 years ago

5 0

Answer:

$x = - 3$

Step-by-step explanation:

The equations given are:

$y_1=6x+10$

$y_2=4x+4$

For the equations to generate the same independent value, then

$y_1=y_2$

This implies that:

$6x+10 = 4x + 4$

Group similar terms to get:

$6x - 4x = 4 - 10$

Simplify to get:

$2x = - 6$

$x = - 3$

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Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

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

IF YOUR SMART PLZ ANSWER!!!

serious [3.7K]

Answer: 3rd jump

Step-by-step explanation:

It needs to be 4.65 to be rounded to 4.7 and the 3rd jump is 4.647

3 0

2 years ago

The rectangle below is dilated by a scale factor of 1/3. Find the perimeter and area of the rectangle below.

Dmitrij [34]

For perimeter you add up the two sides so 15 + 21 equals 36

4 0

2 years ago

Find the area of the parallelogram whose sides are given by vectors a and b. Use the cross product only.

MaRussiya [10]

9514 1404 393

Answer:

√869 ≈ 29.48 square units

Step-by-step explanation:

The vector cross product is ...

(3·1 -5·4, -6·1 +2·4, 6·5 -2·3) = (-17, 2, 24)

The area is the magnitude of this vector, so is ...

√((-17)² +2² +24²) = √(289 +4 +576) = √869 ≈ 29.48

The area is about 29.48 square units.

5 0

2 years ago

How many solutions does this system have?

Alecsey [184]

Answer:

C. infinite

Step-by-step explanation:

Both of the equations overlap so there are infinite solutions to the system of equations.

If this answer is correct, please make me Brainliest!

8 0

3 years ago