Is there a transformation that maps shape I onto shape IV? Explain your answer.

Mathematics

1 answer:

miss Akunina [59]2 years ago

5 0

Answer:

Yes, there is a transformation that maps shape I onto shape IV. Reflecting shape I across the x-axis maps it onto shape IV.

Step-by-step explanation:

edmentum.

You might be interested in

A roller coaster can accommodate 18 riders in 8 minutes. Complete the table with equivalent ratios.

12345 [234]

Answer:

36 riders in 16 minutes, 54 riders in 24 minutes, and 72 riders in 32 minutes.

Step-by-step explanation:

The roller coaster can accommodate 9 riders every 4 minutes. 36 riders in 16 minutes, 54 riders in 24 minutes, and 72 riders in 32 minutes all follow the common ratio.

7 0

2 years ago

A. 5 b. 7 c. 9 d. 11 PLEASE HELPPP I WILL GIVE BRILLIANT!!!

antiseptic1488 [7]

Answer:

B : 7

Step-by-step explanation:

4 0

2 years ago

PLEASE HELP

oee [108]

3:1 ratio meaning for every 3 brown eyed people there are 1 blue eyed. So if you have 12 brown every you have 4 blue eyed. Giving you a total of 16 people.

6 0

2 years ago

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_{ x \to 0 } ~ x^x = 1$

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use Lhopitals rule

 $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$

Therefore,

 $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$

3 0

2 years ago

What is the interquartile range of the following data set?

solniwko [45]

The interquartile range is A. 15.5

5 0

2 years ago