Answer: Choice B. sqrt(2)
Draw out a right triangle in quadrant IV as you see in the attached image below. The horizontal and vertical legs are both 1 unit long. To ensure that the signs are properly set up, I am making the vertical leg BC have a label "-1" to mean this is below the x axis. Note how
tan(theta) = opposite/adjacent = BC/AB = -1/1 = -1
Use the pythagorean theorem to find that the hypotenuse AC is sqrt(2) units long
a^2 + b^2 = c^2
(1)^2 + (1)^2 = c^2
2 = c^2
c^2 = 2
c = sqrt(2)
The secant of theta is the ratio of the hypotenuse over the adjacent side, so we end up with
sec(theta) = hypotenuse/adjacent
sec(theta) = AC/AB
sec(theta) = sqrt(2)/1
sec(theta) = sqrt(2) which is why choice B is the answer
Answer:
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Step-by-step explanation:
To find the slope through 2 points
m = ( y2-y1)/(x2-x1)
= ( -4 - -3)/(3 - 3)
= ( -4+3)/(3-3)
= -1/0
Since it is a number divided by zero, the slope is undefined
Answer:
<h2>Let the common ratio b x thus the part becomes</h2><h2>=5x and 4x respectively</h2>
<h2>Thus,ATQ</h2>
<h2>5x+4x= 54</h2>
<h2>9x=54</h2>
<h2>x= 6</h2>
<h2>Sam will get is £30 (putting the value of x)</h2><h2>and Bethan will get £24</h2>
Answer:
7/3
Explanation:
The slope of the line is equal to the rise/run, or in other words, the number of units the line travels upwards over the number of units the line travels to the right.
We can identify the slope using any two points on the line. Here, we can use the two points that are marked on the picture. The second point is 7 units above the first and 3 units to the right of the first, so the slope of the line is equal to 7/3.
Another way to calculate the slope of the line is the use this formula:
(y2-y1)/(x2-x1)
The first point is at the coordinate (1,-4) and second point is at the coordinate (4,3). When we plug these two coordinates into the equation, we get this:
(y2-y1)/(x2-x1)
->(3-(-4))/(4-1)
When we simply the fraction, we would get 7/3 and that would give us the slope.
I hope this helps!
Answer: $24, $36, $44, $59, $66, $71, $108
Step-by-step explanation:
