Consider isosceles ΔXYZ.
1 answer:
A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The length of the legs XY and YZ is equal to 39 units.
<h3>What is a triangle?</h3>A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angles of a triangle is always equal to 180°. An isosceles triangle has two sides out of three equal.
A.) The value of n can be found by equating XY and XZ together,
XY = XZ
9n + 12 = 15n - 6
12 + 6 = 15n - 9n
18 = 6n
n = 3
Hence, the value of n is 3.
The length of different legs of the triangle is,
B.) XY = 9n + 12 = 9(3) + 12 = 27 + 12 = 39
C.) YZ = 15n - 6 = 15(3) - 6 = 45 - 6 = 39
Hence, the length of the legs XY and YZ is equal to 39 units.
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A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).
Answer:
Step-by-step explanation:
Consider the provided number.
We need to find the approximate value of to the nearest hundredth.
First find two perfect squares that the irrational number falls between.
118 is lying between 100 and 121, therefore the square root value of 118 will be somewhere between 10 and 11.
118 is closer to 121 as compare to 100.
Therefore,
Consider the number
First find two perfect squares that the irrational number falls between.
319 is lying between 289 and 324, therefore the square root value of 319 will be somewhere between 17 and 18.
319 is closer to 324 as compare to 289.
Therefore,