Applying the knowledge of fractions, the bigger one is: one-sixth of 138.
<h3>What are Fractions?</h3>
Fractions can be defined as a part of a digit or number, and are not whole numbers. For example, a third is 1/3, one-sixth is 1/6, which are both fractions.
Therefore, let's evaluate each statement:
1/3 of 45 = 1/3 × 45 = 15
1/6 of 138 = 1/6 × 138 = 23
Therefore, the bigger of the two is: one-sixth of 138.
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The size of the second application given the size of the first application and the expression ( x - 3.45 mb) for the size of the second application is 293.55 MB.
<h3>Equation</h3>
Let
- Size of the first application = x
- Size of the second application= x - 3.45 mb
For instance,
if the size of the first application is 297 MB
Size of the second application= x - 3.45 mb
= 297 MB - 3.45 MB
= 293.55 MB
Therefore, the size of the second application given the size of the first application and the expression for the size of the second application is 293.55 MB
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Answer:
W = 2.5d + 62
Step-by-step explanation:
The calf weighed 62 pounds when they were born. This gives us a base of 62 pounds - the calf cannot weigh less than 62 pounds and it does on day 0. On each day , the calf gains 2.5 pounds. We can times the number to days by 2.5 to get the gain from day 0. We can add these two values together to get the total current weight of the calf.
A method that always works is to find the slope of the given line, then find the negative reciprocal of that. Your result will be the slope of the perpendicular line. Using this slope and the given point, fill in the parameters of the point-slope form of the equation of a line.
For m = slope of given line and (h, k) = given point, the perpendicular line will be
y = (-1/m)(x -h) +k
Often, this equation can be simplified to another appropriate form, such as slope-intercept form (y = mx+b) or standard form (ax+by=c).
_____
The slope of a given line can be found by solving its equation for y. The slope is the coefficient of x in that solution. If the given line is characterized by two points, (x1, y1) and (x2, y2), then its slope is m = (y2-y1)/(x2-x1).
In the unusual case where the given line is vertical (x=<some constant>), the slope of the perpendicular line is zero, and the line you want becomes y=k.
