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loris [4]
3 years ago
14

The vertices of polygon ABCD are at A(1,1), B(2,3), C(3,2), and D(2,1). ABDC is reflected across the x-axis and translated 2 uni

ts up to form polygon A’B’C’D’. Match each vertex of polygon A’B’C’D’ to it’s coordinations
Mathematics
2 answers:
Iteru [2.4K]3 years ago
8 0

Answer:

A

Step-by-step explanation:

-a dilation with a scale factor greater than 1 and then a translation

velikii [3]3 years ago
4 0

Answer:


Step-by-step explanation:


A. (1,1)

B. (2, -1)

C. (3, 0)

D. (2,1)

yes so i have matched the vertices of the polygon

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Substitue the expression 5-3x into the second equation
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Answer:

<h2><u>X= 1</u></h2>

Step-by-step explanation:

<em>y = 5 - 3x</em>

<em>5x - 4y = -3</em>

you substitute for y since you know what y equals.

so, it will then be

5x -4(5 -3x) = -3

you then multiply -4 by everything in the parenthesis

so then it will be

5x - 20 + 12x = -3

Then combine like terms

5x + 12x = 17x

then the new equation is

17x -20 = -3

the next step is to and 20 to both sides since it is negative you do the opposite.

17x - 20 = -3

<u>      + 20     +20</u>

17x = 17 divide 17 by both sides sice 17 is being multiplied by x you do the opposite and divide and you get 1.

<h2><u>so, X = 1</u></h2>
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dem82 [27]

Answer:

The inequality is 6 + 3x ≤ 12 or x ≤ 2 .

Step-by-step explanation:

Given that $6 is for admission which is a fixed amount, $3 is charged per hour and must not spend more than $12. So the inequality will be :

Let x be the no. of hours,

6 + 3x ≤ 12

Solve :

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mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
3 years ago
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