5 squared is 5x5. And 2 cubed is 2x2x2. So, the answer would be 200.
It’s would be 4/25
Mark as brainliest
Answer:
Step-by-step explanation:
hello :
the first graph : 1 and -1 at second ; 2
To prove that <span>ΔABC ≅ ΔMQR using SAS, we show that two sides with the intersection angle are congruent.
From the diagram, it is shown that CA is congruent to RM.
From the first option, given that </span>m∠A = 64° and AB = MQ = 31 cm, then we have CA = RM, AB = MQ, and CAB = RMQ (i.e. m∠A = <span>m∠M = 64°). </span>
This shows that the first option is correct.
From the second option, given that CB = MQ = 29 cm, then we have CA = RM, <span>CB = MQ, but ACB is not congruent to RMQ.
Thus the second option in not correct.
From the third option, </span>m∠Q = 56° and CB ≅ RQ, then we have CA = RM, CB = RQ, ACB = 60<span>°, but we do not know the value of MRQ.
Thus the third option is not correct.
From the fourth option, </span>m∠R = 60° and AB ≅ MQ, then we have <span>CA = RM, AB = MQ, RMQ = </span>64<span>°, but we do not know the value of CAB.
Thus the fourth option is not correct.
From the fifth option</span>, <span>AB = QR = 31 cm, then we have </span><span>CA = RM, </span><span>AB = QR, but we do not know the value of CAB or MRQ.
Thus, the fifth option is not correct.
Therefore, the additional information that </span><span>could be used to prove ΔABC ≅ ΔMQR using SAS is </span><span>m∠A = 64° and AB = MQ = 31 cm</span>
Answer:
A = 175, B = π/6, C = 345
Step-by-step explanation:
The given variation of the volume in the tank and time includes are;
At t = 0 the volume in the tank, y = Maximum volume, 
= 520 gallons
At t = 6 the volume in the tank, y = Minimum volume, 
= 170 gallons
The function that models the situation is y = A·cos(B·x) + C
Given that the function that models the situation is the cosine function, we have;
A = The amplitude = (The maximum - The minimum)/2
∴ A = (520 - 170)/2 = 175
A = 175
The period = The time to change from maximum to minimum = 2 × The time to change from maximum (t = 0) to minimum (t = 6)
∴ The period = 2 × (6 - 0) = 12
The period = 12 seconds = 2·π/B
∴ B = 2·π/12 = π/6
B = π/6
C = The vertical shift = Th minimum + A = (The maximum + The minimum)/2
∴ C = (520 + 170)/2 = 345
C = 345
