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3 years ago

If xy = 1, which is equivalent to x(x – 1)(y + 1)?

Mathematics

1 answer:

dmitriy555 [2]3 years ago

7 0

Answer:

B) x^2 – 1

Step-by-step explanation:

x(x – 1)(y + 1)

FOIL (x-1)(y+1)

firsy: xy

outer 1x

inner -1y

last (1)(-1) =-1

Add them together xy +x-y-1

x( xy +x-y-1)

Replace xy with 1

x( 1+x-y-1)

x(+x-y)

Distribute

x^2-xy

But xy=1

x^2 -1

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A rectangular tank measuring 50 cm x 30 cm x 60 cm was completely filled with water. 1 5 of the water in the tank was transferre

Ierofanga [76]

Answer:

Step-by-step explanation:

Volume of the rectangular tank = Length * Width * Height

Given

Length = 50cm

Width = 30cm

Height = 60cm

Get the volume:

Volume = 50*30*60

Volume = 90,000cm³

Hence the volume of the rectangular tank is 90,000cm³.

If 1/5 of the volume of water was transferred to another tank, the volume transferred will be:

Amount transferred = 1/5 * 90,000

Amount transferred = 18000cm³

Amount left in the tank = 90,000 - 18,000

<em>Amount of water left in the tank = 72,000cm³</em>

5 0

3 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$