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igomit [66]
2 years ago
15

Trapezoid W′X′Y′Z′ ​ is the image of trapezoid WXYZ under a dilation through point C.

Mathematics
2 answers:
dem82 [27]2 years ago
8 0

Answer:

Option C.

Step-by-step explanation:

Given information: WX=8m and W'X'=6m

It is given that Trapezoid W′X′Y′Z′ ​ is the image of trapezoid WXYZ under a dilation through point C.

Scale factor is the ratio of corresponding sides of image and preimage.

\text{Scale factor}=\frac{Corresponding side of image}{Corresponding side of preimage}

For the given figure,

\text{Scale factor}=\frac{W'X'}{WX}

\text{Scale factor}=\frac{6}{8}

\text{Scale factor}=\frac{3}{4}

Since image and preimage lie on the same side of the center of dilation. So, the scale factor is positive number.

Therefore, the correct option is C.

Luda [366]2 years ago
4 0

The answer is 3 over 4

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There was no snow on the ground when it started falling at midnight at a constant rate of 1.5 inches per hour. At
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Answer:

The required piece-wise linear function is

\left\{\begin{matrix}1.5x & 0\leq x

Step-by-step explanation:

Consider the provided information.

Let x represents the number of hours and S(x) represents the depth of snow.

There was no snow on the ground when it started falling at midnight at a constant rate of 1.5 inches per.

That means the depth of snow will be:

S(x)=1.5x\ \ \ 0\leq x< 4

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From mid night to 4:00 a.m the depth of snow will be 1.5×4=6 inches.

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S(x)=3(x-4)+6\ \ \ 4\leq x

7:00 a.m. to 9:00 a.m., snow was  falling at a constant rate of 2 inches per hour.

From mid night to 7:00 a.m the depth of snow will be 3(7-4)+6=15 inches.

If we want to calculate the total depth of snow from midnight to 9:00 am we need to add 15 inches in 2(x-7) where 7≤x≤9

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3 years ago
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Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

  1. f : X \rightarrow Y is surjective implies there exists a function h : Y \rightarrow X such that  f\circ h = 1_Y.
  2. If there exists a function h : Y \rightarrow X such that  f\circ h = 1_Y, then f : X \rightarrow Y is surjective

Let us start by the first implication.

Our hypothesis is that the function f : X \rightarrow Y is surjective. From this we know that for every y\in Y there exist, at least, one x\in X such that y=f(x).

Now, define the sets X_y = \{x\in X: y=f(x)\}. Notice that the set X_y is the pre-image of the element y. Also, from the fact that f is a function we deduce that X_{y_1}\cap X_{y_2}=\emptyset, and because  f the sets X_y are no empty.

From each set X_y  choose only one element x_y, and notice that f(x_y)=y.

So, we can define the function h:Y\rightarrow X as h(y)=x_y. It is no difficult to conclude that f\circ h(y) = f(x_y)=y. With this we have that f\circ h=1_Y, and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function  h:Y\rightarrow X  such that f\circ h=1_Y.

Take an element y\in Y, then f\circ h(y)=y. Now, write x=h(y) and notice that x\in X. Also, with this we have that f(x)=y.

So, for every element y\in Y we have found that an element x\in X (recall that x=h(y)) such that y=f(x), which is equivalent to the fact that f is surjective. Therefore, the prove is complete.

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