For this case, what we must do is find three times the perimeter of the figure.
The length of the arch is given by:
Where,
r: radius of the circle
alpha: central angle in radians
We have then:
Then, the perimeter of the figure is:
Answer:
If David runs around the field three times, he covers a distance of 648 yards.
Answer:
4th statement is true.
Step-by-step explanation:
We have been two box plots, which represents the number of calories in each snack pack of crackers and cookies. We are asked to find the correct statement about our given box plots.
1. More packets of crackers have 80 calories than any other number of calories.
We can see that median of box plot representing calories of cookies is 80. This means that half of the packets of crackers have less than 80 calories and half of the packets have more than 80 calories, therefore, 1st statement is false.
2. The value 70 is an outlier for the number of calories in the cookie pack.
Since an outlier is 1.5 times the interquartile range.
Since any number less than 67.5 will be an outlier and 70 is grater than 67.5, therefore, 70 is not an outlier in number of calories in cookie packs and 2nd statement is false.
3. The upper quartile of the cookie data is equivalent to the maximum in the cracker data.
We can see that upper quartile of cookie data is 105 and the maximum in cracker data is 100. Since 105 is greater than 100, therefore, 3rd statement is false.
4. The number of calories in each pack of cookies has a greater variation than the number of calories in each pack of crackers.
Since range and IQR are good measures of variation of box-plots, so we will find the range and IQR of our both box-plots.
We have already seen that IQR of cookie packs is 15.
We can see that the range of calories in cookie packs (45) is greater than range of calories in cracker packs (30) and IQR of calories in cookie packs (15) is greater than IQR of calories in cracker packs (10), therefore, 4th statement is true.
Answer:
The simplest form is x/(x + 3)
Step-by-step explanation:
* To simplify the rational Expression lets revise the factorization
of the quadratic expression
* To factor a quadratic in the form x² ± bx ± c:
- First look at the c term
# If the c term is a positive number, and its factors are r and s they
will have the same sign and their sum is b.
# If the c term is a negative number, then either r or s will be negative
but not both and their difference is b.
- Second look at the b term.
# If the c term is positive and the b term is positive, then both r and
s are positive.
Ex: x² + 5x + 6 = (x + 3)(x + 2)
# If the c term is positive and the b term is negative, then both r and s
are negative.
Ex: x² - 5x + 6 = (x -3)(x - 2)
# If the c term is negative and the b term is positive, then the factor
that is positive will have the greater absolute value. That is, if
|r| > |s|, then r is positive and s is negative.
Ex: x² + 5x - 6 = (x + 6)(x - 1)
# If the c term is negative and the b term is negative, then the factor
that is negative will have the greater absolute value. That is, if
|r| > |s|, then r is negative and s is positive.
Ex: x² - 5x - 6 = (x - 6)(x + 1)
* Now lets solve the problem
- We have two fractions over each other
- Lets simplify the numerator
∵ The numerator is
- Factorize its denominator
∵ The denominator = x² + 2x - 3
- The last term is negative then the two brackets have different signs
∵ 3 = 3 × 1
∵ 3 - 1 = 2
∵ The middle term is +ve
∴ -3 = 3 × -1 ⇒ the greatest is +ve
∴ x² + 2x - 3 = (x + 3)(x - 1)
∴ The numerator =
- Lets simplify the denominator
∵ The denominator is
- Factorize its denominator
∵ The denominator = x² - 2x
- Take x as a common factor and divide each term by x
∵ x² ÷ x = x
∵ -x ÷ x = -1
∴ x² - 2x = x(x - 1)
∴ The denominator =
* Now lets write the fraction as a division
∴ The fraction = ÷
- Change the sign of division and reverse the fraction after it
∴ The fraction =
* Now we can cancel the bracket (x + 2) up with same bracket down
and cancel bracket (x - 1) up with same bracket down
∴ The simplest form =