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3 years ago

Which sequence of transformations does not preserve distance and angle? rotation and reflection

Mathematics

2 answers:

Serggg [28]3 years ago

7 0

Answer:

C. Rotation and dilation.

Step-by-step explanation:

We are asked to choose the sequence of transformations that does not preserve distance and angle from our given choices.

Since we know that preserving distance means if the length of a segment is 5 units, then it will be 5 units after the transformation. Preserving angle means that if the measure of an angle is x degrees, then it will be x degrees after the transformation as well.

The transformations that preserve distance and angle are known as rigid transformations. Translation, rotation and reflection are rigid transformation, while dilation is non rigid transformation.

Therefore, rotation and dilation will not preserve distance and angle as dilation enlarges or compresses a figure.

Vanyuwa [196]3 years ago

6 0

Answer: rotation and dilation.

While rotation only rotates (obviously) the figure, dilation preserves the shape of the figure (that is, preserves the angles) but changes the distances: it makes the figure smaller.

The other 2 sequences of transformations only change the position of the figure, but the angles and distances remain intact.

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Evaluate the following <img src="https://tex.z-dn.net/?f=%20-%20%20%5Csqrt%5B4%5D%7B256%7D%20" id="TexFormula1" title=" -

poizon [28]

Ur answer is here
Hope it is helpful to u

6 0

3 years ago

Pleaseee someone help me :( Ive been stuck on this for an hour

Whitepunk [10]

Answer:

The number of times organism B's population is larger than organism A's population after 8 days is 32 times

Step-by-step explanation:

The population of organism A doubles every day, geometrically as follows

a, a·r, a·r²

Where;

r = 2

The population after 5 days, is therefore;

Pₐ₅ = = 32·a

The virus cuts the population in half for three days as follows;

The first of ta·2⁵ he three days = 32/2 = 16·a

The second of the three days = 16/2 = 8·a

After the third day, Pₐ = 8/2 = 8·a

The population growth of organism B is the same as the initial growth of organism A, therefore, the population, P₈ of organism B after 8 days is given as follows;

P₈ = a·2⁸ = 256·a

Therefore, the number of times organism B's population is larger than organism A's population after 8 days is P₈/Pₐ = 256·a/8·a = 32 times

Which gives, the number of times organism B's population is larger than organism A's population after 8 days is 32 times.

Hope this helps you out! :)

Also, it's OK! This was actually pretty hard to figure out!

6 0

3 years ago

What is the distance between -3.5 and 12.4

Komok [63]

Answer:

We have 12.4 - ( - 3.5 ) = 12.4 + 3.5 = 15.9 is the distance between 2 points.

Step-by-step explanation:

4 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$