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2 years ago

Which rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C'?

Mathematics

1 answer:

cluponka [151]2 years ago

7 0

The rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C' is (x, y) → (x + 7, y - 6).

<h3>What is translation?</h3>

When a line is translated, it means the line is moved from one position to another. The coordinate for triangle ABC are A(-3 ,4) , B(-4,1) and C(-2,1).

The coordinate of triangle A'B'C' is A'(4,-2), B'(3,-5), and C'(5,-5).

From above, it can be seen that the image A'B'C' is obtained from the pre-image ABC by translating the vertices of the image by 7 units to the right and 6 units down.

Therefore, the rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C' is (x, y) → (x + 7, y - 6)

Hence, the rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C' is (x, y) → (x + 7, y - 6).

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Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.