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Oksana_A [137]
2 years ago
9

AB has a vértices at (-5,1) and (-3,7). After a transformation, the image had endpoints at (3,-1) and (5,5). Describe the transf

ormation
Mathematics
2 answers:
Dafna11 [192]2 years ago
7 0

If the image is A'B' we have, for the point A x=-5 goes to x = 3 ( a translation)

Its a translation of 8 to the right.

The same goes for point B the x coordinate goes 8 to the right.

The y coordinate for A translates 1 to -1 That is 2 units down

Same goes for B coordinate of y

Translation is 8 units to right and 2 down.

Scrat [10]2 years ago
5 0
That looks like a translation; let's check.  We have

A(-5,1), B(-3,7), A'(3,-1), B'(5,5)

If it's a translation by T(x,y) we'd have

A' = A + T

B' = B + T

so 

T = A' - A = (3,-1) - (-5,1) = (8,-2)

and also 

T = B' - B = (5, 5) - (-3, 7) = (8,-2)

They're the same so we've verified this transformation is a translation by (8,-2), eight right, two down.

 



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frosja888 [35]

Answer:

x²+11x+28

Step-by-step explanation:

(f - g) (x) = f(x) - g(x)

x² + 12x + 32 - (x + 4) = x² + 11x + 28

6 0
2 years ago
In the circle below, chord AB passes through the center of circle 0. If the radius OC is perpendicular to chord AB and has lengt
Airida [17]

Answer:

Option (C)

Step-by-step explanation:

In the given circle, AB is a chord passing through the center O.

Therefore,chord AB is a diameter of the circle and OC is the radius.

AO = OB = OC

OC⊥AB, and length of OC = 7 cm

By applying Pythagoras theorem in right ΔBOC,

BC² = OC² + OB²

BC = \sqrt{OC^{2}+OB^{2}}

     = \sqrt{7^2+7^2}

     = 7\sqrt{2}

     ≈ 9.9 units

Therefore, Option (C) will be the answer.  

7 0
3 years ago
Can you help me plz <br><br> take your time tho
AnnZ [28]

Answer:

d

Step-by-step explanation:

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7 0
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Which answer best describes the complex zeros of the polynomial function?
marysya [2.9K]

<u>Answer-</u>

<em>D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.</em>

<u>Solution-</u>

The given polynomial is,

f(x)=x^3-x^2+6x-6

The zeros of the polynomials are,

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\Rightarrow x^3-x^2+6x-6=0

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\Rightarrow (x^2+6)(x-1)=0

\Rightarrow (x^2+6)=0,(x-1)=0

\Rightarrow x=\sqrt{-6},\ x=1

\Rightarrow x=1,\ \pm \sqrt6i

Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1

3 0
3 years ago
Use the system 4x -2y=21, y - 2x =10. Write each equation in slope intercept form. Compare the slopes and y-intercepts. Make a c
Korvikt [17]

Answer:

its no solutions

Step-by-step explanation:

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