There are 5 letters in the word "prime"
Imagine we had 5 slots to fill. They are empty initially.
Slot 1 has 5 choices to pick from
Once we pick a letter, we have 4 choices left over for slot 2
Slot 3 will have 3 choices
Slot 4 will have 2 choices
Slot 5 will have 1 choice
We have this countdown: 5,4,3,2,1
which multiplies out to 5*4*3*2*1 = 120
There are 120 unique ways to arrange the letters. Order matters. Because order matters, this is a permutation.
You changed the mixed numeral into a fraction correctly.
Now you need to multiply the fractions.
Multiply the numerators together, and multiply the denominators together.
Now we could multiply out the numerator and denominator, and then we'd need to reduce the fraction, but before we multiply, we see there is a factor of 5 both in the numerator and in the denominator, so we can reduce before we multiply.
Answer:
<u>Perimeter</u>:
= 58 m (approximate)
= 58.2066 or 58.21 m (exact)
<u>Area:</u>
= 208 m² (approximate)
= 210.0006 or 210 m² (exact)
Step-by-step explanation:
Given the following dimensions of a rectangle:
length (L) = meters
width (W) = meters
The formula for solving the perimeter of a rectangle is:
P = 2(L + W) or 2L + 2W
The formula for solving the area of a rectangle is:
A = L × W
<h2>Approximate Forms:</h2>
In order to determine the approximate perimeter, we must determine the perfect square that is close to the given dimensions.
13² = 169
14² = 196
15² = 225
16² = 256
Among the perfect squares provided, 16² = 256 is close to 252 (inside the given radical for the length), and 13² = 169 (inside the given radical for the width). We can use these values to approximate the perimeter and the area of the rectangle.
P = 2(L + W)
P = 2(13 + 16)
P = 58 m (approximate)
A = L × W
A = 13 × 16
A = 208 m² (approximate)
<h2>Exact Forms:</h2>
L = meters = 15.8745 meters
W = meters = 13.2288 meters
P = 2(L + W)
P = 2(15.8745 + 13.2288)
P = 2(29.1033)
P = 58.2066 or 58.21 m
A = L × W
A = 15.8745 × 13.2288
A = 210.0006 or 210 m²
Answer:
(1;-2)
Step-by-step explanation:
x=-2y-3
2(2y+3)=-6y-9-5
10y=-20
y=-2
x=4-3=1
Answer:
Part A: a histogram
Step-by-step explanation:
Part A:
A histogram would be the best to use since they gave values that are x - y, or a ranged amount for the score.
Part B:
You would set up each value for the scores on the bottom. You would make the number of students on the left in increments of 1 (1,2,3,4,5 etc. )
You would make the first value ( 0-4 ) go up to 4 students. The second value ( 5-9) would go to 5. The third value ( 10-14 ) would go to 2. The fourth and fifth values ( 15-19 and 20-24 ) would go to 3.