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Gnoma [55]
4 years ago
11

HELP ME!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Makovka662 [10]4 years ago
8 0
You go to connexus?! Hold on lemme check my notes! :)
Irina18 [472]4 years ago
7 0
Regular.Hope that helped
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What is 9060000000 In standard form
kolezko [41]

Answer:

9.06 × 10

Step-by-step explanation:

5 0
3 years ago
Solve and show the steps to the picture attached
lina2011 [118]

Answer:

The image of point (2, 4) after transformation is (-2, 13) ⇒ D

Step-by-step explanation:

Let us revise some transformation

Translation right-left OR up-down

  • If the function f(x) translated horizontally to the right by h units, then its image is g(x) = f(x - h)  ⇒ add x-coordinate of every point by h
  • If the function f(x) translated horizontally to the left by h units, then its image is g(x) = f(x + h)   ⇒ subtract x-coordinate of every point by h
  • If the function f(x) translated vertically up by k units, then its image is g(x) = f(x) + k  ⇒ add y-coordinate of every point by k
  • If the function f(x) translated vertically down by k units, then its image is g(x) = f(x) - k  ⇒ subtract y-coordinate of every point by k

A vertical stretching and a vertical compression (or shrinking)

  • If m > 1, the graph of y = m • f(x) is the graph of f(x) vertically stretched by factor m.  ⇒ multiply every y-coordinate by m
  • If 0 < m < 1 (a fraction), the graph of y = m • f(x) is the graph of f(x) vertically shrunk (or compressed) by factor m  ⇒ multiply every y-coordinate by m

Let us use these rules to solve the question.

∵ Point (2, 4) lies on y = x²

∵ The image of y = x² is y = 3(x + 4)²+1

→ The rules of transformation is y = m(x + h)² + k

∵ m is the scale factor of a vertical stretched

∵ m = 3

→ By using the 5th rule above

∴ The y-coordinate of point (2, 4) should multiply by 3

∴ The image of the point is (2, 4 × 3) = (2, 12)

∵ h is the horizontal translation to the left

∵ h = 4

→ By using the 2nd rule above

∴ The x-coordinate of point (2, 12) should subtract by 4

∴ The image of the point is (2 - 4, 12) = (-2, 12)

∵ k is the vertical translation up

∵ k = 1

→ By using the 3rd rule above

∴ The y-coordinate of point (-2, 12) should add by 1

∴ The image of the point is (-2, 12 + 1) = (-2, 13)

∴ The image of point (2, 4) after transformation is (-2, 13)

3 0
3 years ago
A fair coin is tossed three times. What is the probability of obtaining one Head and two Tails?(A fair coin is one that is not l
emmainna [20.7K]
3/8 is the answer ;)
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3 years ago
An ant on the ground is 22 feet from the foot of a tree. She looks up at a bird at the top of the tree, and her line of sight ma
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6 0
4 years ago
Let f(x)=2−|4x−2|. Show that there is no value of c such that f(3)−f(0)=f'(c)(3−0). Why does this not contradict the Mean Value
xenn [34]

Answer:

The mean value theorem is valid if f(x) is continuous in the interval (0,3) and differentiable in the interval (0,3), the problem is that f(x)=2−|4x−2| is not differentiable in x = 1/2 because 4*1/2 - 2 = 0 and the function |x| is not differentiable in x = 0.

f'(x) = (-4)*(4x−2)/|4x−2|

f(3) = 2−|4*3−2| = 8

f(0) =  2−|4*0−2| = 0

Replacing in f'(c) = f(3)−f(0)/(3−0)

(-4)*(4c−2)/|4c−2| = (8 - 0)/3

(-4)*(4c−2)*3/8 = |4c−2|

-3/2*4c + 3/2*2 = |4c−2|

-6c + 3 = |4c−2|

That gives us two options

-6c + 3 = 4c−2

5 =  10c

1/2 = c

or

6c - 3 = 4c−2

-1 = -2c

1/2 = c

But f'(1/2) is not defined, therefore there is  no value of c such that f(3)−f(0)=f'(c)(3−0).

3 0
3 years ago
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