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2 years ago

11

Help? Is this true or false? Ill give brainliest

2 answers:

Tatiana [17]2 years ago

8 0

It would be false i think

Vlad [161]2 years ago

5 0

Answer:

False I believe the answer is false

Step-by-step explanation:

The slope is 6 but the y - intercept is 2

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A square is made up of an L-shaped region and three transformations of the region. If the perimeter of the square is 40 units, w

Mila [183]

Answer:

20 units

Step-by-step explanation:

This implies that the square can be divided into four equal L-shaped regions. These regions with respect to transformation forms a square.

Perimeter of the square is 40 units. Since a square has equal length of sides, thus each side of the square is 10 units.

Thus, each L-shape region has dimensions; 8 units, 5 units, 5 units and 2 units.

Perimeter of each L-shape region = the addition of the length of each side of the shape

Perimeter of each L-shape region = 8 + 5 + 5 + 2

= 20 units

3 0

3 years ago

Giraffes are the tallest land animals.A male giraffe can grow as tall as 6 yards.How tall would the giraffe be in feet?

Katyanochek1 [597]

20 feet im not sure i hope this helps

3 0

3 years ago

Read 2 more answers

two friends bought a $14 pizza. Each person bought their own drink. The total cost of the food can be represented by the express

yanalaym [24]

The expression might be x=7

4 0

2 years ago

mihalych1998 [28]

Answer:

$=4.+2.$

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector $\vec b$ is a linear combination of $\vec a_1$ and $\vec a_2$ if there are two scalars $x_1, x_2$ such as

$\vec b=x_1\vec a_1+x_2\vec a_2$

In our case, all the vectors are given in $R^3$ but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

$\vec a_1=,\ \vec a2=,\ \vec b=$

We set the equation

$=x_1.+x_2.$

Multiplying both scalars by the vectors

$=+$

Equating each coordinate, we get

$4x_1-4x_2=8$

$5x_1+3x_2=26$

$-4x_1+3x_2=-10$

Adding the first and the third equations:

$-x_2=-2$

$x_2=2$

Replacing in the first equation

$4x_1-4(2)=8$

$4x_1=8+8$

$x_1=4$

We must test if those values make the second equation become an identity

$5(4)+3(2)=20+6=26$

The second equation complies with the values of $x_1$ and $x_2$ , so the solution is

$=4.+2.$

8 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