5. Which lengths could represent the side lengths of a right triangle? 1.5, 15. 230
II. 6, 10, 136
III. 7, 13, 229
A I, II and III
B. II only
C. I and II only
D. None of the above
6. Three students draw triangle with the side lengths shown. All three s
triangle. Which of the students are incorrect? What mistake might they
PLS HELP

What is the number 321,000,000 in scientific notation? also what is the number? ?

Cone w has a radius of 8 cm and a height of 5 cm. square pyramid x has the same base area and height as cone w. paul and manuel disagree on how the volumes of cone w and square pyramid x are related. examine their arguments. which statement explains whose argument is correct and why? paul manuel the volume of square pyramid x is equal to the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is three times the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is (area of base)(h) = (200.96)(5) = 1,004.8 cm3. paul's argument is correct; manuel used the incorrect formula to find the volume of square pyramid x. paul's argument is correct; manuel used the incorrect base area to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect formula to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect base area to find the volume of square pyramid x.

Which sum or difference identity would you use to verify that cos (180° - q) = -cos q? a.) sin (a -b) = sin a cos b – cos a sin b b.) cos (a -b) = cos a cos b – sin a sin b c.) cos (a -b) = cos a cosb + sin a sin b d.) sin (a + b) = sin a cos b + cos a sin b