Watch one is the right one

Write the square of the binomial as a polynomial: (-1 1/3p + 6q)^2

Thepotemich [5.8K]

Answer:

$\frac{16}{9}$ p² -16pq + 36q²

Step-by-step explanation:

Given

(- $\frac{4}{3}$ p + 6q)²

= (- $\frac{4}{3}$ p + 6q)(- $\frac{4}{3}$ p + 6q)

Each term in the second factor is multiplied by each term in the first factor, that is

- $\frac{4}{3}$ p(- $\frac{4}{3}$ p + 6q) + 6q(- $\frac{4}{3}$ p + 6q)

= $\frac{16}{9}$ p² - 8pq - 8pq + 36q² ← collect like terms

= $\frac{16}{9}$ p² - 16pq + 36q²

5 0

4 years ago

8 k+d use k=2 and d=3

sveta [45]

8×2=16

16+d

d=3

16+3=19

5 0

3 years ago

Given constraints: x is greater than or equal to 0, y greater than or equal to 0, 2x + 2y is greater than or equal to 4, x + y i

Alik [6]

Graph the inequalities given by the set of constraints. Find points where the boundary lines intersect to form a polygon. Substitute the coordinates of each point into the objective function and find the one that results in the largest value.

6 0

3 years ago

Read 2 more answers

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

Can someone do this for me please?! Just the answer pls

Lana71 [14]

<h2>For the number 1..</h2>

<h2>angle at center = 2×angle at circumference</h2><h2>the angle at center (o) is equal to 141</h2>

therefore:

<h2>141 = 2x</h2><h2>divide both sides by 2</h2><h2>x = 70.5</h2>

option (A).

( the number 2 and 3 questions aren't correct )

6 0

2 years ago