Answer:
Step-by-step explanation: the slope would be 1 and the y intercept would be 4.
Answer:
Step-by-step explanation:
You are to assume in both problems that the two triangles are similar. That is a very dangerous assumption -- especially in later math classes. But in this case, there is no other way to do the problem. The two sets of triangles look like they are proportional. So set up two ratios that are = to each other
On the left
long side small triange / long side large triangle = base small triangle / base large triangle
On the right
hypotenuse/longest leg = hypotenuse / longest leg.
Problem A
Set up the similar Proportion
5/40 = x/20 Cross Multiply
40x = 20*5
40x = 100 Divide by 40
x = 100/40
x = 2.5
Problem B
Again set up the similar triangle proportion
15/10 = x/2 Multiply by 2
15*2/10 = x
3 = x
<span>Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral.
Let's see now, Delmar has a line segment WX and has drawn 2 circles whose radius is the length of WX, centered upon W and centered upon X. Sounds to me that all he needs to do is select one of the intersections of those 2 circles and use that at the 3rd point of the equilateral triangle and draw a line from that point to W and another line from that point to X. Doesn't matter which of the two intersections he chooses, just needs to pick one. Looking at the available options, only the 1st one which is "Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral." matches my description, so that is the correct choice. The other choices tend to do rather bizarre things like create a perpendicular bisector of WX and for some unknown reason, claim that bisector is somehow a side of a desired equilateral triangle.</span>
Answer:obtuse
Step-by-step explanation:
Well, we can make it 3 + 5
3 + 5 = 8
Hope I helped!
~ Zoe
