Plot the point (-7, -5). We are in quadrant 3.
We also know that tan θ = opp/adj.
Cot θ = adj/opp.
Let us use a^2 + b^2 = r^2 to show you how to find r, the radius aka hypotenuse.
Look: (-7)^2 + (-5)^2 = r^2
If you should ever need to find r, do this:
(-7)^2 + (-5)^2 = r^2
(49) + 25) = r^2
74 = r^2
Take the square root on both sides of the equation to find r.
sqrt{74} = sqrt{r^2}
sqrt{74} = r
It is ok to simplify the sqrt{74} but not needed.
We now have the three sides of the triangle that is form in quadrant 3.
We can now read cot θ from the triangle itself.
So, cot θ = adj/opp = (-7)/(-5) or 7/5.
No need to find r but I simply wanted to show you how it's done in case you are given a question where r must be found.
Answer:
The correct option is;
line a and line b
Step-by-step explanation:
Lina a passes through (-1, 0) and (1, 2)
Line b passes through (2, 2) and (-1, -1)
Line c passes through (0, 3) and (1, 0)
Line d passes through y = 1
An equation that has no solution
By examination, from the plot of the coordinates using an online tool, the system that would have no solution is the system of line a and line b because they are parallel lines and cannot meet or have a point where there is a common solution, therefore, there are no solutions when the equations of the two lines are combined in one equation
Slope of line a = -2/-2 = 1
Slope of line b = 3/3 = 1
The slopes of lines a and b are equal, therefore, line a and line b are parallel.
#7
2,4,8,16
15,30,60,120
#9
1,2,3,
4,8,12
Step-by-step explanation:
Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.
Type ni xi si
Graded 42 0.486 0.187
No-fines 42 0.359 0.158
To find - a. Formulate the above in terms of a hypothesis testing problem.
b. Give the test statistic and its reference distribution (under the null hypothesis).
c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.
Proof -
a.)
Hypothesis testing problem :
H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
b)
Test statistic :




⇒Z(cal) = 3.3687
Z(tab) = 1.96
As Z (cal) > Z(tab)
So, we reject H0 at 5% Level of significance
p-value = 0.99962
Hence
There is significant difference in mean conductivity at the two materials.