Explain how you know that 70 is greater than 1/3 but less than three 2/3
1 answer:
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Answer:
The answer is (C) 8
Step-by-step explanation:
First, let's calculate the length of the side of the square.
, where
is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:
which for
and
transforms as
. The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then,
,
and 10 are a Pythagorean triple. From this,
or
while
or
. This leads us with the set of coordinates:
and
. (A)
The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):
Let's do this for the point (6, 8).
The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation , however, we only need to find its slope in order to find a perpendicular line to it. Thus,
Then, a perpendicular line has an slope (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.
(1)
(2)
This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope based on the perpendicularity condition. Thus, we can form the system:
(1)
(2)
with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:
- (0, 0), (6, 8), (14, 2), (8, -6)
- (0, 0), (8, 6), (14, -2), (6, -8)
- (0, 0), (-6, 8), (-14, 2), (-8, -6)
- (0, 0), (-8, 6), (-14, -2), (-6, -8)
Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution and
. By combining this points we get the following squares:
- (0, 0), (10, 0), (10, 10), (0, 10)
- (0, 0), (0, 10), (-10, 10), (-10, 0)
- (0, 0), (-10, 0), (-10, -10), (0, -10)
- (0, 0), (0, -10), (-10, -10), (10, 0)
See the attached second attached figure. Therefore, 8 squares can be drawn
In this question , we have to simplify the given ratio, which is
12xy:3x
First we have to see which factor is common in numerator and denominator,
We can write 12 as 3 times 4, that is
So the common factor is 3x.
In the next step, we cancel out 3x
And that's the required simplified form .
First, we should know exponential formula, for example exp (LnA) =A, for all A>0, and as for logarythm, LnA^p=pLnA, for all A>0, and for all value of p
so y=3^x=exp[Ln(3^x)]=exp(xLn3), let's search the graph of f(x)=exp(xLn3) f is defined in R, and its teminals limits are 0, and + infinity, f' (x) = Ln3.exp(xLn3), it is positif for all value of x, the function is strictly increasing, f(0)=Ln3, please look at the figure 1
the same method with y=2(4.5)^x, we find y=exp[Ln (2(4.5)^x)]=
exp[Ln (2) +Ln(4.5)^x)], because (LnAB=LnA + LnB), so y=2.exp[Ln(4.5)^x)]=2.exp(x[Ln(4.5)]), which domain is R, nd its teminals limits are 0, and + infinity, f' (x)=2.Ln(4.5) exp(x[Ln(4.5)]), which is positif for all value of x, the function is strictly increasing,and then f(0)=2Ln(4.5)
please look the figure 2
so y=3^x=exp[Ln(3^x)]=exp(xLn3), let's search the graph of f(x)=exp(xLn3) f is defined in R, and its teminals limits are 0, and + infinity, f' (x) = Ln3.exp(xLn3), it is positif for all value of x, the function is strictly increasing, f(0)=Ln3, please look at the figure 1
the same method with y=2(4.5)^x, we find y=exp[Ln (2(4.5)^x)]=
exp[Ln (2) +Ln(4.5)^x)], because (LnAB=LnA + LnB), so y=2.exp[Ln(4.5)^x)]=2.exp(x[Ln(4.5)]), which domain is R, nd its teminals limits are 0, and + infinity, f' (x)=2.Ln(4.5) exp(x[Ln(4.5)]), which is positif for all value of x, the function is strictly increasing,and then f(0)=2Ln(4.5)
please look the figure 2