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lozanna [386]
3 years ago
9

Which data point is an outlier in this graph?

Mathematics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

I can't find the answer.

Step-by-step explanation:

You should put a photo of the graph and I would be able to tell :)

Sorry

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The expression (4c − 3d)(3c + d) is equivalent to:
Umnica [9.8K]

(4c - 3d)(3c + d) =

= 12c² + 4cd - 9cd - 3d² =

= <u>12c² - 5cd - 3d²</u>

4 0
3 years ago
rotation of 90 degrees counterclockwise about the origin, point O, then a reflection across the x-axis reflection across the y-a
SpyIntel [72]

Answer:

The point O can be (x,y)  or (-x,y) or (-x,-y) or ( x,-y) reason being that you haven't given point O lies in which quadrant.

Rotation through 90° counter clockwise

(x,y) = (-y,x)

(-x,y)=(-y,-x)

(-x,-y)=(y,-x)

(x,-y) =(y,x)

Then Reflection across X axis has taken place.

(-y,x) = (-y,-x)

(-y,-x)=(-y,x)

(y,-x)=(y,x)

(y,x)=(y,-x)

Then a reflection across the y-axis has taken place.

(-y,-x)=(y,-x)

(-y,x) = (y,x)

(y,x) =(-y,x)

(y,-x)=(-y,-x)

Then a translation a units to the right and b units up has taken place.

(y,-x)=(y+a,-x+b)

(y,x) =(y+a, x+b)

(-y,x) =(-y+a,x+b)

(-y,-x)=(-y+a,-x+b)

Then a rotation of 180 degrees counterclockwise about the origin has taken place.

(y+a,-x+b)=[-(y+a),-(-x+b)]

(y+a, x+b)=[- (y+a), -(x+b)]

(-y+a,x+b)=[-(-y+a),-(x+b)]

(-y+a,-x+b) =[-(-y+a),-(-x+b)]

Now Again a reflection across the Y axis has taken place.

[-(y+a),-(-x+b)]=[(y+a),-(-x+b)]

[-(y+a),-(x+b)]=[(y+a),-(-x+b)]

[-(-y+a),-(x+b)]=[(-y+a),-(x+b)]

[-(-y+a),-(-x+b)]=[(-y+a),-(-x+b)]

Totally depends on value of a and b on which quadrant these point lies.



7 0
3 years ago
Can anyone solve this question please
levacccp [35]

Answer:

Step-by-step explanation:

by definition : The inverse of a relation consisting of points of the form (x,y) is the set of points (y,x)

if (1,7) is a point belong to f(x), then the inverse has to be (7,1) not (8,1)

3 0
3 years ago
GIVING BRAINLIEST PLEASE HELP
BabaBlast [244]

Answer:

S=16×2

S=32

................................

3 0
3 years ago
Read 2 more answers
Thank you for anyone who knows this
katen-ka-za [31]

Answer:  

  • Part (a) Vertex = (-3, -1)
  • Part (b) Vertex = (4, 5)
  • Part (c) function 2; max value = 5

=====================================================

Explanations:

Part (a)

In this case, the vertex is the highest point, so the vertex of function 1 is (-3, -1)

If the parabola was flipped so that it opened upward, instead of downward, then the vertex would be the lowest point.

----------------------------

Part (b)

We could graph this function to see where the highest or lowest point is, but there's another way we could find the vertex using algebra.

The given equation is in the form y = ax^2 + bx + c where in this case

  • a = -2
  • b = 16
  • c = -27

The a and b values are plugged into the formula below to get

h = -b/(2a)

h = -16/(2*(-2))

h = -16/(-4)

h = 4

This is the x coordinate the vertex since the vertex is (h,k)

To find k, we plug that value of h into the original function to find its corresponding y value.

y = -2x^2 + 16x - 27

y = -2(4)^2 + 16(4) - 27

y = -2(16) + 16(4) - 27

y = -32 + 64 - 27

y = 5

We found that h = 4 leads to k = 5. The vertex is (h,k) = (4, 5)

----------------------------

Part (c)

We'll compare the y values of each vertex

  • Function 1 has a vertex of (-3, 1)
  • Function 2 has a vertex of (4, 5)

The vertex with the larger y value is the winner, so that would be function 2. This parabola reaches a higher peak compared to the other curve.

The max value here is y = 5. In other words, it's the maximum output the function can produce.

If you graphed functions 1 and 2 together on the same xy grid (see below), you'll see that the curve for function 2 reaches a higher peak.

7 0
2 years ago
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