Answer:
C) Distributive Property
Step-by-step explanation:
You're <em>distributing</em> in the 3 when you multiply it by the 5 and the 2 in the algebraic process:
➦ ➦
3•(5 + 2) = (3 • 5) + (3 • 2)
So, the 3 was <em>distributed</em> between the 5 and 2 when you multiplied.
Hence, this is the distributive property.
Is there a picture we can look at ?
The area of a rectangle can be calculated by the product of its length and the width. Here we are given the area and the length so we can simply calculate for width. The perimeter is the sum of all the side lengths of the shape. We calculate as follows:
Area = lxw
1/6 = 1.5w
w = 1/9
Perimeter = 2l + 2w
Perimeter = 5/9
Hope this answers the question. Have a nice day.
The percentage of tardiness among the 102 pupils is 2.94.
<h2>Calculation of the percentage</h2>
According to the query,
Course A: Total number of students = 102
Number of tardy students = 3
Percent of students tardy in course A = (number of tardy students/total
number of students) * 100
=(3/102)*100
= 2.941
≈2.94
In case of course B: Total number of students = 85 & tardy student = 5
So, the percent of tardy student = (5/85)*100 = 5.88 %
Therefore, it is concluded that the percentage of student tardiness in course A is 2.94.
Learn more about the percentage here:
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<span>First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.
You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...
The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span>
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<span> where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.</span>