-6i+10i answer please

Which of the following systems of linear inequalities is represented by the

algol [13]

Answer:

Option D is correct.

Step-by-step explanation:

Option D is correct.

y < 2 and y >= x

Because the dotted line shows that y < 2 and y >=x is shown is represented by diagonal line.

5 0

3 years ago

Read 2 more answers

Help please due now!

MissTica

Answer:

8

Step-by-step explanation:

3(2)-2(1-2) / 3-2

6-2(-1) / 1

6 + 2 / 1

8 / 1

= 8

6 0

3 years ago

Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6 into the quadratic formula?

ZanzabumX [31]

Answer:

shows correct substitution of the values a, b, and c from the given quadratic equation into quadratic formula.

Step-by-step explanation:

Given: The quadratic equation

We have to show the correct substitution of the values a, b, and c from the given quadratic equation into quadratic formula.

The standard form of quadratic equation is then the solution of quadratic equation using quadratic formula is given as

Consider the given quadratic equation

Comparing with general quadratic equation, we have

a = -3 , b = -2 , c = 6

Substitute in quadratic formula, we get,

Simplify, we have,

Thus, and x_{2}=\frac{2-\sqrt{76} }{-6}

Simplify, we get,

Thus, shows correct substitution of the values a, b, and c from the given quadratic equation into quadratic formula.

3 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

15 POINTS PLEASE HELP ASAP!!<br> (pls dont write random answers)

quester [9]

Answer:

its either d or a

Step-by-step explanation:

3 0

3 years ago