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2 years ago

what are the coordinates of the image after a transformation of (x,y)-->(x-3,y+7) ? A(2,5) B(6,-10) C( -12, 15)

Mathematics

1 answer:

ivanzaharov [21]2 years ago

7 0

Applying the translation rule, the coordinates are:

A'(-1,12).
B'(3, -3).
C'(-15, 22).

The translation rule is:

$(x,y) \rightarrow (x - 3, y + 7)$

Thus, for the image, the transformation is applied for each coordinate. Then:

$A' = (2 - 3, 5 + 7) = (-1, 12)$

$B' = (6 - 3, -10 + 7) = (3, -3)$

$C' = (-12 - 3, 15 + 7) = (-15, 22)$

A similar problem is given at brainly.com/question/24721131

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Distance between (-6, -2) and (0,-1)

s344n2d4d5 [400]

Hey there! I'm happy to help!

To find the distance between two points, you square the difference of the x-values and square the difference of the y-values, add them, and then you square root it!

First, we'll add our two x-values.

-6-0= -6

We square it, which means to multiply it by itself.

-6(-6)=36 (If you multiply an even number of negative numbers, your answer is positive. Since we have two negative numbers, we get positive 36)

Now, we do the same with the y-values.

-2-(-1)=-1 (two negatives make it a plus, as in minus minus 1 is plus one.)

We square it.

-1(-1)=1

Now, we add these x and y value differences.

36+1=37

Now, we find the square root using a calculator.

√37≈6.08

Have a wonderful day!

5 0

3 years ago

grace and her friends were to roll the two dice 100 times, how many of the 100 times could they expect to roll the same number o

LenaWriter [7]

Answer:

The probability of both dice having the same number is 636, as there are 36 different outcomes, 6 of which have two of the same number, i.e. (1,1),(2,2),....

The expected number of rolls of this type in 100 pairs of dice rolls is 100∗636

4 0

2 years ago

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$