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2 years ago

Solve the following equation: 3 x one half

Mathematics

1 answer:

Kamila [148]2 years ago

6 0

<h3>hello!</h3>

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Let's simplify the expression:-

$\bigstar{\boxed{\frac{3}{1} *\frac{1}{2}}$

Multiply 3 times 1 and 1 times 2:-

$\bigstar{\boxed{\pmb{\frac{3}{2} }}$

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Hope everything is clear; if you need any clarification/explanation, kindly let me know, and I'll comment and/or edit my answer :)

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The coordinates of the vertices of triangle ABC are A(3,6), B(6, 3), and C(9, 9). If triangle ABC Is dilated by 1/3x and 1/3y to

katovenus [111]

Answer:

$A' = (1,2)$

$B'=(2,1)$

$C' =(3,3)$

Step-by-step explanation:

Given

$A = (3,6)$

$B = (6,3)$

$C = (9,9)$

$Scale = \frac{1}{3}$

Required

Determine the vertices of ABC

If truly the question asks for the vertices of ABC, then the vertices are:

$A = (3,6)$ $B = (6,3)$ $C = (9,9)$

However, I'm quite sure that's not the requirement of the question. So, I'll solve for A'B'C'

To do this, we simply multiply the vertices of ABC by the scale of dilation.

$(A'B'C') = Scale\ Factor * (ABC)$

$A' = \frac{1}{3} * A$

$A' = \frac{1}{3} * (3,6)$

$A' = (\frac{3}{3},\frac{6}{3})$

$A' = (1,2)$

$B' = \frac{1}{3} * B$

$B' = \frac{1}{3} * (6,3)$

$B' = (\frac{6}{3},\frac{3}{3})$

$B'=(2,1)$

$C' = \frac{1}{3} * C$

$C' = \frac{1}{3} * (9,9)$

$C' = (\frac{9}{3},\frac{9}{3})$

$C' =(3,3)$

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3 years ago