Answer:
A. 121 ⇒ III. 11
B. 64 ⇒ II. 4 and IV. 8
C. 27 ⇒ I. 3
D. 125 ⇒ V. 5
E. 16 ⇒ II. 4
Step-by-step explanation:
Let us find the correct answer
∵ 121 = 11 × 11
∴ The square root 121 is 11
∴ A. 121 ⇒ III. 11
∵ 64 = 8 × 8
∴ The square root of 64 is 8
∵ 64 = 4 × 4 × 4
∴ The cube root of 64 is 4
∴ B. 64 ⇒ II. 4 and IV. 8
∵ 27 = 3 × 3 × 3
∴ The cube root of 27 is 3
∴ C. 27 ⇒ I. 3
∵ 125 = 5 × 5 × 5
∴ The cube root of 125 is 5
∴ D. 125 ⇒ V. 5
∵ 16 = 4 × 4
∴ The square root of 16 is 4
∴ E. 16 ⇒ II. 4
Answer:
see the explanation
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠PQS+m∠SQR=m∠PQR ----> equation A (by Addition Angle Postulate)
we have that
m∠PQR=90° ----> equation B given problem (because is a right angle)
substitute equation B in equation A
m∠PQS+m∠SQR=90°
Remember that
Two angles re complementary is their sum is equal to 90 degrees (Definition of complementary angles)
therefore
m∠PQS and m∠SQR are complementary angles
Answer:
bru its 132
Step-by-step explanation:
(x²+5)½ + (x²+5)³/²
->(x+√5) +(x³+√125)
->(x+√5)+(x³+5√5)
->x+x³+√5+5√5
->2x³+6√5
->2(x³+3√5)