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klio [65]
3 years ago
7

Shape 1 and shape 2 are plotted on a coordinate plane. Which rigid transformation can you perform on shape 2 to show that shape

2 is congruent to shape 1?
a 45° rotation clockwise about the origin

a translation 6 units to the left and 8 units up

a reflection across the x-axis

a translation 3 units to the left and 6 units up

Mathematics
2 answers:
FinnZ [79.3K]3 years ago
8 0
I hope this helps you



vazorg [7]3 years ago
6 0
According to given information, <span>a translation 6 units to the left and 8 units up for shape 2 would show that it is congruent to shape 1.

So, option B is your answer.

Hope this helps!
</span>
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What is the GCF of 8x^3y^2+20x^2y^4?
DochEvi [55]

Answer:

B) 4x^2y^2

Step-by-step explanation:

8x^3y^2+20x^2y^4

4(2x^3y^2+5x^2y^4)

4x^2(2xy^2+5y^4)

4x^2y^2(2x+5y^2)

So the answer is B

5 0
3 years ago
The equation 4x2 – 24x + 4y2 + 72y = 76 is equivalent to
DerKrebs [107]

Answer:

Option 4 is correct.

The equation 4x^2 -24x + 4y^2 + 72y = 76 is equivalent to 4(x-3)^2 + 4(y+9)^2 =436

Step-by-step explanation:'

Given equation: 4x^2 -24x + 4y^2 + 72y = 76

First group the terms with x and those with y;

(4x^2-24x)+(4y^2+72y) = 76

Next, we complete the squares.

We can do this by adding a third term such that the x terms and the y terms are perfect squares.

For this we must either add the same value on the other side of the equation or subtract the same value on the same side so that the equality is maintained.

⇒4(x^2-6x) +4(y+18y) = 76

or

4(x^2 -6x +3^2 -3^2) + 4(y^2 +18y +9^2 -9^2) = 76

4(x^2-6x + 3^2) - 36 + 4(y^2+18y +9^2) - 324 = 76

4(x-3)^2 + 4(y+9)^2 - 360 =76

Add 360 on both sides we get;

4(x-3)^2 + 4(y+9)^2 =360 +76

Simplify:

4(x-3)^2 + 4(y+9)^2 =436

Therefore, the given equation is equivalent to 4(x-3)^2 + 4(y+9)^2 =436

5 0
3 years ago
Line segments A prime B prime and A B. The distance from the center of dilation P to A prime is 3. The distance from P to A is 5
OverLord2011 [107]
The answer is P to A prime is 3 I think

Good luck!
8 0
2 years ago
What number must you add to complete the square?<br> x^2 + 2x = -1
Andru [333]
\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}


If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248


\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}

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