Answer:
obtuse, acute
Step-by-step explanation:
edg 2020
Let's say
. Let's find point
so that we can find
.
- Leah walks 40 yards south.

- Leah walks 60 yards west.

- Leah walks 10 yards north.

- Leah walks 20 yards east.

We have found that
.
Now think about this scenario visually. We started at the center of something, which we call point
, and then started moving around until we got to point
. We can then form line
between the points. However, realize that we can actually make a triangle. Just think of one of the legs as part of the x-axis and the other leg as part of the y-axis.
We can find the length of these parts, which is simply the absolute value of the coordinates of point
. It may be a little hard to think about, but essentially, we can form a triangle with sides that consist of part of the x-axis, part of the y-axis, and
. We also know that the lengths of the legs are 40 and 30.
Since we are given the two lengths of the legs on the triangle and trying to find the length of the hypotenuse, we can use the Pythagorean Theorem. This states:

and
are the lengths two legs of the triangle
is the length of the hypotenuse
Thus, substituting in our values, we find:


The length of
is 50.
Answer:
x = 6
Step-by-step explanation:
This shows that the angle being split is 90 degrees.
<em>We know that 32 degrees are already known so we can do </em><em>90 - 32 = 58</em>
Now we know that 58 is the angle of 9x + 4
9x + 4 = 58
Subtract 4 from both sides
9x = 54
Divide both sides by 9 to isolate x
x = 6
Answer: rectangles are congruent if both of them have the opposite sides are equal
Step-by-step explanation: rectangles are congruent if both of them have the opposite sides are equal. Two squares are congruent if both of them have the same edges.
Answer:
0.79
Step-by-step explanation:
Here,
Let X be the event that the flights depart on time
Let Y be the event that flights arrive on time
So,
X∩Y will denote the event that the flights departing on time also arrive on time.
Let P be the probability
P(X∩Y)=0.65
And
P(X)=0.82
We have to find P((Y│X)
We know that
P((Y│X)=P(X∩Y)/P(X) )
=0.65/0.82
=0.79
So the probability that a flight that departs on schedule also arrives on schedule is: 0.79