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Vedmedyk [2.9K]

1 year ago

15

How can I make sure that each point of a Triangle was reflected across the line?

1 answer:

larisa86 [58]1 year ago

4 0

For example, for A and A'

Let:

R=(0,0) the reference point:

$\begin{gathered} \text{Distance between A and R:} \\ d=\sqrt[]{(4-0)^2+(2-0)^2}=2\sqrt[]{5} \\ \text{Distance between A' and R} \\ d=\sqrt[]{(4-0)^2+(-2-0)^2}=2\sqrt[]{5} \end{gathered}$

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Edg vector operations, any help appreciated!

viktelen [127]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: Add \:\: -6 \hat i - 6\hat j \:\:with \:\; Vector \:\; c$

____________________________________

$\large \tt Solution \: :$

Vector d can be represented as :

$\qquad \tt \rightarrow \: - 2 \hat i - 2 \hat j$

Vector c can be represented as :

$\qquad \tt \rightarrow \: 4 \hat i + 4\hat j$

we have to create vector d from vector c

So, let's assume a vector x, such that sum of vector x and vector c equals to vector d

$\qquad \tt \rightarrow \: x + ( 4 \hat i + 4 \hat j) = - 2 \hat i - 2 \hat j$

$\qquad \tt \rightarrow \: x = - ( 4 \hat i + 4 \hat j) - 2 \hat i - 2 \hat j$

$\qquad \tt \rightarrow \: x = (- 4 \hat i - 2 \hat i) + ( - 4 \hat j - 2 \hat j)$

$\qquad \tt \rightarrow \: x = - 6 \hat i -6 \hat j$

Henceforth, in order to get vector d, we need to add (-6i - 6j) in vector c

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

8 0

2 years ago

Please help asap! Due tonight!

Contact [7]

Answer:

A, $\sqrt{17}$

Step-by-step explanation:

a^2 + b^2 = c^2

8^2 + b^2 = 9^2

64 + b^2 = 81

b^2 = 17

b = $\sqrt{17}$

4 0

3 years ago

Read 2 more answers

NEED HELP ASAP!!!!! PLEASE HELP ME

mina [271]

ANSWER

X = -9

EXPLANATION: you subtract 180 - 70 giving you 110

Equation set up: x + 49 + x + 79 = 110

7 0

3 years ago

Read 2 more answers

(03.05 MC) Solve the rational equation x divided by 2 equals x squared divided by quantity x minus 2 end quantity, and check for

ycow [4]

Answer:

x = 0 and x = -2 are solutions of the given rational equation.

Step-by-step explanation:

We must solve the following rational equation:

$\frac{x}{2} = \frac{x^{2}}{x-2}$

Now we present the procedure:

1) $\frac{x}{2} = \frac{x^{2}}{x-2}$ Given

2) $x\cdot (x-2) = 2\cdot x^{2}$ Compatibility with multiplication/Existence of the multiplicative inverse/Definition of division/Modulative property.

3) $x^{2}-2\cdot x = 2\cdot x^{2}$ Distributive property/ $a^{b}\cdot a^{c} = a^{b+c}$

4) $x^{2} + 2\cdot x = 0$ Compatibility with addition/Existence of the additive inverse/Modulative property/Reflexive property

5) $x \cdot (x+2) = 0$ Distributive property/ $a^{b}\cdot a^{c} = a^{b+c}$

6) $x = 0\, \lor\, x = -2$ Result

Now we check the rational equation with each root:

x = 0

$\frac{x}{2} = \frac{x^{2}}{x-2}$

$\frac{0}{2} = \frac{0^{2}}{0-2}$

$0 = \frac{0}{-2}$

$0 = 0$

x = 0 is a solution of the rational equation.

x = -2

$\frac{x}{2} = \frac{x^{2}}{x-2}$

$\frac{-2}{2} = \frac{(-2)^{2}}{-2-2}$

$-1 = -1$

x = -2 is a solution of the rational equation.

4 0

3 years ago

tatuchka [14]

750 international minutes

100-25=75

75÷0.10=750

3 0

3 years ago