The only geometric objects which can be defined using the "locus of points" idea is a straight line, circle, and angle bisector.

What is the length of the hypotenuse of the triangle 12ft 35 ft

Sunny_sXe [5.5K]

Answer: 37

Step-by-step explanation: a“2”+b“2”=c“2”

12“2”+35“2”=c“2”

24+1225=c“2”

1369=c“2”

sqrt 1369=c

c=37

5 0

4 years ago

PLEASE I NEED IT NOW.....In order to solve for the variable in the equation 1-(x+2)+2x = 5(2x-5) - X, Mikel first applies the di

loris [4]

Answer:

1 -x -2 +2x = 10x -25 -x

Step-by-step explanation:

1-(x+2)+2x = 5(2x-5) - X,

Distribute

1 -x -2 +2x = 10x -25 -x

3 0

4 years ago

Two boats are traveling from different cities, with two different destinations. Boat 1 is traveling from City A (West 120 degree

-Dominant- [34]

It would be c I did all the calculations for it

8 0

3 years ago

Juan is trying to find a secure place to keep his important papers. He has insurance policies and the titles to his car and hous

lesantik [10]

Answer:

Juan should put his papers in a place that nobody else knows excise. Like in the chimney, the secret door or the atti.

Step-by-step explanation:

you can’t just put it somewhere people see or know

5 0

3 years ago

A track star runs two races on a certain day. The probability thathe wins the first race is 0.7, the probability that he wins th

NARA [144]

Answer:

a) 80% probability that he wins at least one race.

b) 30% probability that he wins exactly one race.

c) 20% probability that he wins neither race.

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 he wins the first race.

B is the probability that he wins the second race.

C is the probability that he does not win any of these races.

We have that:

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

In which a is the probability that he wins the first race but not the second and $A \cap B$ is the probability that he wins both these races.

By the same logic, we have that:

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

The probability that he wins both races is 0.5.

This means that $A \cap B = 0.5$

The probability that he wins the second race is 0.6

This means that $B = 0.6$

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

$0.6 = b + 0.5$

$b = 0.1$

The probability that he wins the first race is 0.7.

This means that $A = 0.6$

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

$0.7 = a + 0.5$

$a = 0.2$

A) he wins at least one race.

This is

$P = a + b + (A \cap B) = 0.2 + 0.1 + 0.5 = 0.8$

There is an 80% probability that he wins at least one race.

B) he wins exactly one race.

This is

$P = a + b = 0.2 + 0.1 = 0.3$

There is a 30% probability that he wins exactly one race.

C) he wins neither race

Either he wins at least one race, or he wins neither. The sum of these probabilities is 100%.

From a), we have that there is an 80% probability that he wins at least one race.

So there is a 100-80 = 20% probability that he wins neither race.

6 0

3 years ago