Michael and Tyler both ran a half marathon Michael finished in 1 hour 42 minutes and 13 seconds Tyler finished in 97 minutes and 49 seconds who was faster how much faster was he?​

part a: triangles gcb and epb

part b. sides cg and rp are parallel by definition of parallelogram.

angles cgb and e are congruent by being alternate interior angles of parallel lines cut by a transversal. angles gbc and ebp are congruent because they are vertical angles. by aa similarity, the triangles are similar.

part c. since the triangles are similar, the ratios of the lengths of corresponding sides are equal.

calculation of be:

gb/be = cb/bp

(400 ft)/be = 300/200

(400 ft)/be = 3/2

3be = 2 * 400 ft

3be = 800 ft

be = 800/3 ft

calculation of pe:

gc/pe = cb/bp

(350 ft)/pe = 300/200

(350 ft)/pe = 3/2

3pe = 2 * 350 ft

3pe = 700 ft

pe = 700/3 ft

graph #4 (last graph) depict the function   f(x)= -m(x-d)^1/5 i

step-by-step explanation:

we know that

m=slope

but in this function slope is -m that means slope is decreasing and it is decreasing by 1/5 points down the x-axis.

which is shown in graph four

a decreasing slope down x-axis by 1/5 points.

hence 4th graph is the right option for this.

  b:   ∠x = 28.6°, ∠y = 31.2°, ∠z = 120.2°

step-by-step explanation:

this is one of those questions where "test taking skill" is all you need. the only answer that has angles that sum to 180° is the one shown above. the remaining choices do not describe the angles of a triangle.

if you want to do more work than simply adding the offered numbers, you can also check them against the law of sines. it should be true that

  x/sin(x) = y/sin(y) = z/sin(z)

and all of these ratios must be greater than 15, the longest side length.

__

if you want to solve the triangle from scratch, i'd suggest solving for the largest angle using the law of cosines.

  z² = x² +y² -2xy·cos(z)

  z = arccos((x² +y² -z²)/(2xy)) = arccos(-75.11/149.4) = 120.182°

now you can find y from the law of sines

  y = arcsin(y/z·sin(z)) = 31.242°

hello,

elba writes the following proof for the theorem: if the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram:

elba's proof

1. for triangles aob and cod, angle 1 is equal to angle 2 as they are vertical angles.

2. ao = oc and bo = od because it is given that diagonals bisect each other.

3. therefore, the triangles aob and cod are congruent by sas postulate.

4. similarly, triangles aod and cob are congruent.

5. by cpctc, angle abd is equal to angle cdb and angle adb is equal to angle cbd.

6. as the are congruent, the opposite sides of quadrilateral abcd are parallel.

7. therefore, abcd is a parallelogram.

which is the missing phrase in elba's proof?

answer - opposite