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Sidana [21]
2 years ago
14

ΔMNO was dilated, then _____________, to create ΔYHQ.

Mathematics
2 answers:
lesya [120]2 years ago
8 0

Rotated

<h2>Explanation:</h2>

Remember you have to write complete questions in order to find good and exact answer. So in this case I'll provide a general explanation and say that the blank can be filled with the word rotated. First of all, when dilating objects we reduce or enlarge the size of them. It's important to know the scale factor because this tells use whether the object will be larger or smaller, so if we call k the scale factor, then:

  • If k > 0 then the object will be larger than the original.
  • If k < 0 then the object will be smaller than the original.

Moreover, the center of dilation is the point where all the lines that meet corresponding points of the original and dilated figure.

On the other hand, to rotate a shape we turn it about a fixed point that we call the center of rotation. So, in this case we'll have that ΔMNO is similar to  ΔYHQ.

<h2>Learn more:</h2>

Dilation:

#LearnWithBrainly

seropon [69]2 years ago
3 0

Answer:

the Anwser is A

Step-by-step explanation:

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Answer:

h=7

Step-by-step explanation:

We want to solve the equation:

\frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{h^2-25}

We multiply through by the LCM: h^2-25=(h+5)(h-5)

(h^2-25*\frac{1}{h-5}+(h^2-25)*\frac{2}{h+5}=\frac{16}{h^2-25}   \times(h^2-25)

Simplify to get:

h+5+2(h-5)=16

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h+5+2h-10=16

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2 years ago
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olga_2 [115]
From the information given,

3 pounds = $18

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Which arrangement shows −275 , −4.1 , −535 , and −3 in order from least to greatest?
vodomira [7]

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-535, -275, -4.1, -3

Step-by-step explanation:

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1 year ago
Let e1= 1 0 and e2= 0 1 ​, y1= 4 5 ​, and y2= −2 7 ​, and let​ T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps
Furkat [3]

Answer:

The image of \left[\begin{array}{c}4&-4\end{array}\right] through T is \left[\begin{array}{c}24&-8\end{array}\right]

Step-by-step explanation:

We know that T: IR^{2}  → IR^{2} is a linear transformation that maps e_{1} into y_{1} ⇒

T(e_{1})=y_{1}

And also maps e_{2} into y_{2}  ⇒

T(e_{2})=y_{2}

We need to find the image of the vector \left[\begin{array}{c}4&-4\end{array}\right]

We know that exists a matrix A from IR^{2x2} (because of how T was defined) such that :

T(x)=Ax for all x ∈ IR^{2}

We can find the matrix A by applying T to a base of the domain (IR^{2}).

Notice that we have that data :

B_{IR^{2}}= {e_{1},e_{2}}

Being B_{IR^{2}} the cannonic base of IR^{2}

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]   (Data of the problem)

T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]   (Data of the problem)

Writing the matrix A :

A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]

Now with the matrix A we can find the image of \left[\begin{array}{c}4&-4\\\end{array}\right] such as :

T(x)=Ax ⇒

T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]

We found out that the image of \left[\begin{array}{c}4&-4\end{array}\right] through T is the vector \left[\begin{array}{c}24&-8\end{array}\right]

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