Answer:
Step-by-step explanation:
1) A similar triangle must have the same ratios of sides. Triangle ABC and triangle LNM are similar because all the sides of LNM are 1/2 of the sides of ABC. As they are all in a 1:2 ratio, the triangles are similar.
2) ADE and ABC are similar because each angle is the same. Angle D and Angle B are the same because they are shown as the same with the same dot symbolization. Angle E and C are the same because they are shown as the same with the same curvy angle representation. ADE and ABC share A as the same angle, so all three angles are the same.
5) As triangles ABC and EDC are similar, the sides must have the same ratio. ED is similar to AB, EC is similar to AC, and CD and BC are similar. First we find the ratio. The only pair of similar lines are EC and AC, so we have to find the ratio from those two lines. EC is 15 and AC is 10/3, so we do 15 divided by 10/3 and we get 9/2. Therefore, the ratio is 9/2. Now we can find x and y. We can first look at ED and AB. Ed is 12 and AB is y. We know that it must have the same ratio of 9/2, so 12/y must equal 9/2. We can do proportions. 12/y = 9/2, 9y = 24 (you multiply the numbers diagonally). So, we get y = 8/3. Now we can get x the same way. BC and CD are similar, and so x/5 must equal 9/2 as well. We use proportions again and we have x/5 = 9/2, so 45 = 2x and x = 45/2.
Answer:
2a: (c)
5o: (1, 3) and (1,1)
3a: (b)
1a: (d)
4o: (b)
Step-by-step explanation:
2a: the equation of a circle circumference needs to be transformable to the form 
where 
is the center and <em>r</em> is the radius. (a) and (d) can’t be it because they contain non-zero factors on <em>xy</em>. (b) isn’t an equation.
5o: just put the given (<em>x</em>, <em>y</em>) into the equations and see if it holds. (2, 3) isn’t on the circumference of (1) because 
, (3, 1) isn’t on it either because 
.
3a: calculate the value of the left-hand side term of the equation using (<em>x</em>, <em>y</em>) from the given point <em>M</em>. That’s the difference of square distance to the center to the square radius 
. Thus it’s 0 if the point is on the circumference, negative if inside and positive if outside. You get 
, positive, so it’s outside the circle.
1a: see definition from 2a. Here, 
.
4o: insert the y from the straight line equation (r) (which can be equivalently transformed to 
) into the circumference equation. If it yields no solution, that’s outside, it there’s exactly one solution, that’s a tangent and if there are two solutions, it’s a secant. 
There are two solutions, so it’s a secant.
The value of x is -3, or x= -3
