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Pie
3 years ago
15

Anybody want to help me in math please please i will really appreciate it

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0
What do you need help with
You might be interested in
Someone please help me with this​
Sedbober [7]

Answer:

The area of the shaded figure is:

  • <u>20 units^2</u>

Step-by-step explanation:

To obtain the area of the shaded figure, first, you must calculate this as a rectangle, with the measurements: wide (4 units), and long (6 units):

  • Area of a rectangle = long * wide
  • Area of a rectangle = 6 * 4
  • Area of a rectangle = 24 units^2

How the figure isn't a rectangle, you must subtract the triangle on the top, so, now we calculate the area of that triangle with measurements: wide (4 units), and height (2 units):

  • Area of a triangle = \frac{wide*height}{2}
  • Area of a triangle = \frac{4*2}{2}
  • Area of a triangle =\frac{8}{2}
  • Area of a triangle = 4 units^2

In the end, you subtract the area of the triangle to the area of the rectangle, to obtain the area of the shaded figure:

  • Area of the shaded figure = Area of the rectangle - Area of the triangle
  • Area of the shaded figure = 24 units^2 - 4 units^2
  • <u>Area of the shaded figure = 20 units^2</u>

I use the name "units" because the exercise doesn't say if they are feet, inches, or another, but you can replace this in case you need it.

5 0
3 years ago
What is the absolute value of 2 and -2?
xeze [42]

If you mean |2-2| then the absolute value is 0

If you are talking about |2| |-2| then the absolute values are 2 and 2

6 0
3 years ago
A line passes through 2, -1 and 4, 5 what is the equation
SVEN [57.7K]

y1 - y2 / x1 - x2

-1 - 5 / 2 - 4

-6/-2

3

y = mx + b

-1 = 3(2) + b

-1 = 6 + b

-7 = b

y = 3x - 7

Hope this helps! ;)

8 0
3 years ago
Read 2 more answers
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
1) A set of numbers in parentheses separated by a comma like (3,-4) is called an
Kruka [31]
The answer is b. Ordered pairs
5 0
3 years ago
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