All categories

2 years ago

What are the coordinates of the image rectangle for a scale factor of 3 if the center of dilation is the origin?

Mathematics

1 answer:

Ksju [112]2 years ago

7 0

Answer: Choice B)

Explanation:

You mentioned there isn't a diagram to go with this, but I'm assuming that this problem is referring to a previous problem you posted. In that problem, it mentions that point W is at (1,-2). If we apply the scale factor 3, then we're tripling each coordinate. That means x = 1 becomes x = 3, and y = -2 becomes y = -6

We can write it like this:

(1,-2) ---> 3*(1, -2) = (3*1, 3*(-2) ) = (3, -6)

Only choice B has W'(3,-6) so this is likely the final answer.

If we apply this dilation to every point, then that effectively makes quadrilateral W'X'Y'Z' to be three times longer and taller than compared to quadrilateral WXYZ. In other words, its side lengths are 3 times longer.

You might be interested in

William bought some tickets to see his favorite singer. He bought some adult tickets and some children’s tickets, for a total of

Fantom [35]

Set x as adult tickets.
Set y as children's tickets.
x + y = 15
30x + 20y = 270
Solve for x in the first equation.
x + y = 15
x = 15 - y
Plug this into the second equation.
30x + 20y = 270
30(15 - y) + 20y = 270
450 - 30y + 20y = 270
450 - 10y = 270
-10y = -180
y = 18
If there is 18 childrens tickets, there should be -3 adult tickets.
This is impossible, and this impossible answer occured because the question is written wrong.
There are a total of 15 tickets
The smallest costing ticket is the childrens ticket, which costs 20$.
If he only bought children tickets, this would be 20x15 which is 300$.
300$ is over 270$, which makes the question impossible.

8 0

3 years ago

HELP AGAIN!!! Plzzzzz thank you!!!

never [62]

Absolute value of -1/3 is 1/3

6 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

How does a digit in the ten thousand place compere to a digit in the thousand place

Mademuasel [1]

Ten thousand - 10,000

thousand literally just haves 1,000

if the twenty is before the three zeros, it would be 20,000 which is twenty thousand. its the same with other numbers. its super simple!

7 0

3 years ago

What is the ratio of yellow sections to blue sections, in lowest terms?

Sveta_85 [38]

Answer:

5 to 3

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers