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3 years ago

Identify the coordinates of point A after a translation with rule

Mathematics

2 answers:

den301095 [7]3 years ago

8 0

To identify the new coordinates after the translation I need the old coordinates

Bond [772]3 years ago

3 0

Your answers are simple they are (3,2), (2,3), (1,-2), and (1,2). I hope the answers help. ☺

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Complete the multiplication: 2EA

Lady bird [3.3K]

In mathematics, a matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

Given that

E = [ 1 2]

A = | 3 0|

| 2 -1 |

2EA = 2 [1 2] | 3 0|

| 2 -1 |

2EA = [ 14 -4]

8 0

4 years ago

Read 2 more answers

Part 4: Use the information provided to write the vertex formula of each parabola.

sergey [27]

Answer: 1. x = (y - 2)² + 8

$\bold{2.\quad x=-\dfrac{1}{2}(y-10)^2}+1$

3. y = 2(x +9)² + 7

Step-by-step explanation:

Notes: Vertex form is: y =a(x - h)² + k or x =a(y - k)² + h

(h, k) is the vertex
point of vertex is midpoint of focus and directrix: $\dfrac{focus+directrix}{2}$

$\bullet\quad a=\dfrac{1}{4p}$

p is the distance from the vertex to the focus

$focus = \bigg(\dfrac{-31}{4},2\bigg)\qquad directrix: x=\dfrac{-33}{4}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{-31}{4}+\frac{-33}{4}}{2}=\dfrac{\frac{-64}{4}}{2}=\dfrac{-16}{2}=-8\\\\\text{The y-value of the vertex is given by the focus as: 2}\\\\\text{vertex (h, k)}=(-8,2)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{-31}{4}-\dfrac{-32}{4}=\dfrac{1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{4})}=\dfrac{1}{1}=1$

Now, plug in a = 1 and (h, k) = (-8, 2) into the equation x =a(y - k)² + h

x = (y - 2)² + 8

***************************************************************************************

$focus = \bigg(\dfrac{1}{2},10\bigg)\qquad directrix: x=\dfrac{3}{2}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1\\\\\text{The y-value of the vertex is given by the focus as: 10}\\\\\text{vertex (h, k)}=(1,10)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{1}{2}-\dfrac{2}{2}=\dfrac{-1}{2}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}$

Now, plug in a = -1/2 and (h, k) = (1, 10) into the equation x =a(y - k)² + h

$\bold{x=-\dfrac{1}{2}(y-10)^2}+1$

***************************************************************************************

$focus = \bigg(-9,\dfrac{57}{8}\bigg)\qquad directrix: y=\dfrac{55}{8}\\\\\text{Since directrix is y, then the y-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{57}{8}+\frac{55}{8}}{2}=\dfrac{\frac{112}{8}}{2}=\dfrac{14}{2}=7\\\\\text{The x-value of the vertex is given by the focus as: -9}\\\\\text{vertex (h, k)}=(-9,7)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{57}{8}-\dfrac{56}{8}=\dfrac{1}{8}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{8})}=\dfrac{1}{\frac{1}{2}}=2$

Now, plug in a = 2 and (h, k) = (-9, 7) into the equation y =a(x - h)² + k

y = 2(x +9)² + 7

4 0

3 years ago

Round 649,418 to the nearest ten,_____,nearest hundred_____,nearest thousand_____,nearest ten thousand_____,nearest hundred thou

kvasek [131]

649,420
649,420
649,420
650,000
700,000

4 0

3 years ago

Read 2 more answers

Explain how to use the missing pieces strategy to compare two fractions. Include a diagram with your explanation.

kozerog [31]

The missing pieces strategy is another method to compare two fractions. To tell which number is larger, you need to see which one is missing less. For example, you have 3/4 versus 4/5. Each of these fractions needs 1 to complete the whole. You need to subtract the numerator with one to compare them. For 3/4 we have 1/4 and for 4/5 we have 1/5. You know that if you complete 4/5 you will get 4. You know that 5 is greater than four. Therefore, 4/5 is larger than 3/4.

5 0

3 years ago

Write the equation of the line that passes through the points (-6,-1) and (-4,2). Put your answer in fully simplified point-slop

Archy [21]

$(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-6)}}} \implies \cfrac{2 +1}{-4 +6} \implies \cfrac{ 3 }{ 2 }$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{3}{2}}(x-\stackrel{x_1}{(-6)}) \implies {\large \begin{array}{llll} y +1= \cfrac{3}{2} (x +6) \end{array}}$

4 0

1 year ago