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denis23 [38]
3 years ago
14

7 - 2x - 5y=-7 3x + 4y = 7

Mathematics
1 answer:
tino4ka555 [31]3 years ago
4 0
-7 + 7 = 0
-5y = 2x
-5y/5 = y
2x/5 = 2x/5

y = 2x/5

4y = -3x + 7

4y/4 = y

-3x/4 + 7/4

y = -3x/4 + 7/4

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The graph of the equation y=f(x) passes through the points (-3,-5), (-2,-3),(0,1) and (4,9). What is the value of y in the equat
lapo4ka [179]
<h3>Answer:</h3>

-5

<h3>Explanation:</h3>

Each ordered pair is a list of an x-vaue and the corresponding f(x) value.

Hence the first ordered pair listed, (-3, -5), means that when x = -3, f(x) = -5. In other words, f(-3) = -5. By the transitive property of equality,

... y = f(-3) = -5

means

... y = -5

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3 years ago
Which statement is true?
adell [148]

Answer:

Answer is the 2nd one.

Step-by-step explanation:

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3 years ago
What are the coordinates of the vertices of the pre-image given? ry= −x ◦ T1, −2(x, y) 
nataly862011 [7]
<h3><u>Answer:</u></h3>

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

<h3><u>Step-by-step explanation:</u></h3>

We have to find the  coordinates of the vertices of the pre-image given?

Ry= −x ◦ T_1, −2(x, y)

i.e. we have to find the composition of reflection along the line y=-x and translation with the rule:

(x,y) → (x+1,y-2)

Now we have coordinates of A",B",C" and D" as:

A"(-4,-2)

B"(-2,-1)

C"(-2,-4)

D"(-4,-3)

Now we are asked to find the coordinates of the pre-image.

Also when any point is reflected along y=-x then the point is transformed to:

(x,y) → (-y,-x)

Let A,B,C and D are the points of the pre-image.

Hence, the coordinates of the pre-image are given as:

so, the transformation is given as:

A→A'→A"

B→B'→B"

C→C'→C"

D→D'→D"

Where A',B',C',D' represents the transformation after translation and A",B",C",D" represents the transformation after reflection as well.

The coordinates of A' are (2,4)

B' are (1,2)

C' are (4,2)

and D' are (3,4)

Now, the coordinates of pre-image are given as:

A'(x,y)  → A(x-1,y+2)=A(1,6)

B'(x,y)  → B(x-1,y+2)=B(0,4)

C'(x,y)  → C(x-1,y+2)=C(3,4)

and D'(x,y)  → D(x-1,y+2)=D(2,6)

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

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