Find the length of the missing side 15 6

Identify a sequence of transformations that maps triangle ABC onto triangle A"B"C" in the image below.

Murljashka [212]

Answer:

The correct option is B.

Step-by-step explanation:

It is given that triangle A"B"C" is the image of triangle ABC after transformation.

From the given figure it is noticed that the point C lies on positive y-axis and point C" lies on negative x-axis.

It means the figure is rotated either counterclockwise 90° or clockwise 270°. The rotation rule is

$(x,y)\rightarrow (-y,x)$

The corresponding sides of image A"B"C" are smaller than the preimage ABC.

$k=\frac{A"B"}{AB}=\frac{3}{6}=\frac{1}{2}$

Since k<0, therefore the transformation shows the reduction. The dilation rule is

$(x,y)\rightarrow (-\frac{1}{2}y,\frac{1}{2}x)$

$A(-3,0)\rightarrow (-\frac{1}{2}(0),\frac{1}{2}(-3))\rightarrow (0,-1.5)$

$B(3,0)\rightarrow (-\frac{1}{2}(0),\frac{1}{2}(3))\rightarrow (0,1.5)$

$C(0,5.2)\rightarrow (-\frac{1}{2}(5.2),\frac{1}{2}(0))\rightarrow (-2.6,0)$

Option B is correct.

4 0

3 years ago

Read 2 more answers

According to a study conducted in one city, 39% of adults in the city have credit card debts of more than $2000. A simple random

Butoxors [25]

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.39, n = 100$

Then

$s = \sqrt{\frac{0.39*0.61}{100}} = 0.0488$

By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

5 0

3 years ago

Y=1/4x-2 write in standard form

oksian1 [2.3K]

Answer: hvdwc bc3uich3jbf 3ichb3lfbd3piujbvler

ion bro

Step-by-step explanation:

5 0

3 years ago

The height of a seat on a Ferris wheel can be modeled as

Ainat [17]

Answer:

99

Step-by-step explanation:

a p e x

3 0

3 years ago

What number could replace ppp below?

Galina-37 [17]

p = 100

Step-by-step explanation:

$\begin{aligned}\tt\dfrac{20}{p}&=\tt\dfrac{2}{10}\\\tt 2p&=\tt 20\cdot 10\\\tt 2p&=\tt 200\\\tt p&=\tt\dfrac{200}{2}\\\tt p&=\tt 100\end{aligned}$

7 0

2 years ago