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gogolik [260]
3 years ago
11

Given pre-image ABCDE. Which of the transformations resulted in image Point A'?

Mathematics
1 answer:
mihalych1998 [28]3 years ago
8 0
<span>A(x, y) → (x - 3, y + 1)</span>
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How do you find the area of this composite figure?
Illusion [34]
If you notice the picture below

the composite figure is just a trapezoid sitting on top of a rectangle
and then, the rectangle has a triangular hole in it

so.. get the area of the trapezoid   \bf \textit{area of a trapezoid}=A=\cfrac{h}{2}(a+b)\qquad &#10;\begin{cases}&#10;h=height\\&#10;a,b=\textit{parallel sides or bases}&#10;\end{cases}

then get the area of the rectangle, which is just a 12x14
and then get the area of the triangle, which surely you know is 1/2 bh

then, subtract the triangle's area from the rectangle's area

and whatever is left, namely the difference, add that to the area of the trapezoid, and that's the composite's area

namely the area of the trapezoid plus the rectangle's, minus the triangle's

6 0
3 years ago
Combine Like Terms and Evaluate
rosijanka [135]

Answer:

29

Step-by-step explanation:

-2x+4+3y-5x-2y

Combine like terms

-7x+1y+4

evaluate

-7(-3) 1(4) 4

Solve

21+4+4

29

6 0
3 years ago
Read 2 more answers
What is 11,339,800 to the nearest whole number
Bas_tet [7]
11,339,800 is 11,340,000 rounded to the nearest whole number
4 0
3 years ago
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Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh
fredd [130]

With

\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k

we have

\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt

The vector field evaluated over this parameterization is

\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k

so the line integral is

\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt

=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4

6 0
3 years ago
Suppose triangle ABC is reflected over the x-axis. If the distance between point A and A’ is 14, what is the distance between th
andreyandreev [35.5K]

Given:

The triangle ABC is reflected over the x-axis.

The distance between point A and A’ is 14.

To find:

The distance between the x-axis and A’.

Solution:

If a figure is reflected across the x-axis then the corresponding parts are mirror image of each other about the x-axis.

It means the distance between A and x-axis is same as the distance between x-axis and A'.

The distance between point A and A’ is 14.

Let d be the distance between the x-axis and A’. Then,

d+d=14

2d=14

d=\dfrac{14}{2}

d=7

Therefore, the correct option is 1.

7 0
3 years ago
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