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3 years ago

Please Help!! Urgent ( Zoom in to see it better)

Mathematics

1 answer:

ruslelena [56]3 years ago

8 0

1. You are trying to isolate "F" in the equation

$C=\frac{5}{9}(F-32)$ Multiply 9/5 on both sides

$(\frac{9}{5})C=(\frac{9}{5}) \frac{5}{9} (F-32)$

$\frac{9}{5}C=F-32$ Add 32 on both sides

$\frac{9}{5}C+32=F-32+32$

$\frac{9}{5}C+32=F$

THIS IS WRONG

2. You are trying to isolate "y" in the equation

$m=\frac{x+y+z}{3}$ Multiply 3 on both sides

3m = x + y + z Subtract x and z on both sides

3m - x - z = y

THIS IS CORRECT

3. Isolate "r"

$s=\frac{r}{r-1}$ Multiply (r - 1) on both sides

s(r - 1) = r Distribute s into (r - 1)

sr - s = r Subtract sr on both sides

-s = r - sr Factor out r from (r - sr)

-s = r(1 - s) Divide (1 - s) on both sides

$\frac{-s}{1-s}=r$

THIS IS WRONG

4. Isolate "b"

$A=\frac{1}{2}(a+b)$ Multiply 2 on both sides

2A = a + b Subtract a on both sides

2A - a = b

THIS IS CORRECT

5. Isolate "y"

$m=\frac{x+y}{2}$ Multiply 2 on both sides

2m = x + y Subtract x on both sides

2m - x = y

THIS IS WRONG

The 2nd and 4th one is right

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Quadrilateral ABCD with vertices A(0, 6), B(-3, -6), C(-9, -6), and D(-12, -3): a) dilation with scale factor of 1/3 centered at

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a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively.

<h3>How to perform transformations with points</h3>

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

$P(x,y)$ - Original point
$P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ , then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ ,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

$C'(x,y) = (-13, -7)$

$D'(x,y) = (-12,-3)+(-5,-1)$

$D'(x,y) = (-17, -4)$

The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively. $\blacksquare$

To learn more on transformation rules, we kindly invite to check this verified question: brainly.com/question/4801277

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