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o-na [289]
3 years ago
7

The perimeter of an irregular shape with 4

Mathematics
1 answer:
sdas [7]3 years ago
4 0

Answer:

9

Step-by-step explanation:

Three sides are

  • 4cm
  • 2cm
  • 6cm

Let other side be x

\rm \mapsto \: x + 4 + 2 + 6 = 21 \\ \rm \mapsto \: x + 6 + 6 = 21 \\   \rm \mapsto \: x + 12 = 21 \\  \rm \mapsto \: x = 21 - 12 \\  \rm \mapsto \: x = 9

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Solve x2 - 8x + 3 = 0 by completing the square.
suter [353]
The correct answer for this would be the second option. The equation that is used in the process in completing the square is this: <span>(x - 4)2 = 13.
The solution is this:
(x-4)(x-4) = 13
x^2 - 4x - 4x + 16 = 13
x^2 - 8x +16 = 13
x^2 - 8x = 13 -16
x^2 - 8x = -3
x^2 - 8x + 3 = 0
Hope this answers your question. Have a great day!</span>
8 0
3 years ago
Given that f(x)=x^2-3x-54 and g(x)=x+6, find (f•g)(x) and express the result in standard form
love history [14]

Answer:

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Step-by-step explanation:

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7 0
2 years ago
Write the final coordinates of the image of D, E, A after the following transformations
cluponka [151]

Step-by-step explanation:

well, the translation (x+6, y+1) creates the following points :

D goes from (-4,-4) to (2,-3)

E goes from (-4,-2) to (2,-1)

A good from (-1,-4) to (5,-3)

the reflection over the x-axis leaves then the x values unchanged, and negative y values turn positive (or vice versa, if we had a case here).

so,

D' = (2,3)

E' = (2,1)

A' = (5,3)

5 0
2 years ago
Evaluate f(x) = 2x - 4 for x = 5
algol13
Plug in x = 5
f(5) = 2(5) - 4
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Solution: f(5) = 6
8 0
3 years ago
Which graph is generated by this table of values? x –4 0 3 y 1 2 3
Nookie1986 [14]

Answer: See the graph attached.

Step-by-step explanation:

1. To solve this exercise you must plot the points given in the table of the problem, which are shown below:

(-4,1), (0,2), (3,3)

2. You must plot the first value of each ordered pair on the x-axis.

3. You must plot the second value of each ordered pair on the y-axis.

4. Therefore, when you plot them, you obtain the graph shown attached.

7 0
4 years ago
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