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3 years ago

What is the sequence of transformations that maps triangle DEF to triangle D’E’F’?

Mathematics

1 answer:

den301095 [7]3 years ago

4 0

For the first box, you will put "reflect over the y axis". Then in the second box, it would be "translate 2 units to the right".

When we reflect over the y axis, the x coordinate changes from positive to negative. Eg: D = (2,-1) ---> (-2,-1). Then we shift 2 units to the right by adding 2 to the x coordinate, so (-2,-1) ---> (0,-1) which is where point D' is located. Points E and F follow the same pattern.

You might be interested in

A plane averaged 500 mph on a trip going east, but only 200 mph on the return trip. The total flying time in both directions was

adoni [48]

Let's assume time taken by plane on one way distance as 't'

The total flying time in both directions was 10.5 hr

so, time taken in other direction =10.5-t

now, we are given

A plane averaged 500 mph on a trip going east

so, total distance of one way is

$=500t$

total distance of other way is

$=200(10.5-t)$

now, we can that distance is same from both sides

so, we get

$500t=200(10.5-t)$

now, we can solve for t

$500t=2100-200t$

$700t=2100$

$t=3$

now, we can find one way distance

$=500\times 3$

$=1500$ miles..............Answer

6 0

3 years ago

19. Find the current in a circuit if the power is 500 W (watts) and the resistance is 25 ohms. Round off your answer to two deci

HACTEHA [7]

Answer:

Step-by-step explanation:

The formula that relates current, power and resistance is

$I=\sqrt{\frac{P}{R}}$

Where

I is the current (in amperes)

P is the power (in watts)

R is the resistance (in ohms)

We know P = 500 and R = 25, we plug them into the formula and solve for I:

$I=\sqrt{\frac{P}{R}}\\I=\sqrt{\frac{500}{25}}\\I=\sqrt{20}\\ I=4.47$

Correct answer is 4.47 Amperes, or choice A.

4 0

4 years ago

Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$