Well you could use the equation 32=2x or 32=(2*x) but if you're looking for an answer that is other than turning this into an equation, then it would be impossible to find an answer. I hope this helped ^^
Answer/Step-by-step explanation:
Recall: SOHCAHTOA
1. Reference angle = 70°
Adjacent side = x
Hypotenuse = 6 cm
Apply CAH. Thus,
Cos 70 = adj/hyp
Cos 70 = x/6
6 × cos 70 = x
2.05 = x
x = 2.05 cm
2. Reference angle = 45°
Adjacent side = x
Hypotenuse = 1.3 m
Applying CAH, we would have the following ratio:
Cos 45 = adj/hyp
Cos 45 = x/1.3
1.3 × cos 45 = x
0.92 = x
x = 0.92 m
3. The who diagram is not shown well. Some parts are missing, however you can still solve the problem just the same way we solved problem 1 and 2.
For the first line we have a slope of (y2-y1)/(x2-x1)
(2--2)/(1--1)=4/2=2 so we have:
y=2x+b, now solve for b with either of the points, I'll use: (1,2)
2=2(1)+b
b=0 so the first line is:
y=2x
Now the second line:
(1-10)/(4--2)=-9/6=-3/2 so far then we have:
y=-3x/2+b, using point (4,1) we solve for b...
1=-3(4)/2+b
1=-6+b
b=7 so
y=-3x/2+7 or more neatly...
y=(-3x+14)/2
...
The solution occurs when both the x and y coordinates for each are equal, so we can say y=y, and use our two line equations...
2x=(-3x+14)/2
4x=-3x+14
7x=14
x=2, and using y=2x we see that:
y=2(2)=4, so the solution occurs at the point:
(2,4)
