The function f is continuous on the closed interval [0,6] and has valu

1.	The function f is continuous on the closed interval [0,6] and has values given in the table above. The trapezoidal approximation found with 3 subintervals is 52. What is the value of k?❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ
Answer» Answer: k = 10 Step-by-step explanation: The formula for the AREA of a trapezoid is 1/2(b_1 + b_2)h where h = DELTA (X) Therefore, A = 1/2(4+k)2 + 1/2(k+8)2 + 1/2(8+12)2 = 54 Solving that equation YIELDS k = 10.

The function f is continuous on the closed interval [0,6] and has values given in the table above. The trapezoidal approximation found with 3 subintervals is 52. What is the value of k?❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ

Answer»

Answer:

k = 10

Step-by-step explanation:

The formula for the AREA of a trapezoid is 1/2(b_1 + b_2)h where h = DELTA (X)

Therefore, A = 1/2(4+k)2 + 1/2(k+8)2 + 1/2(8+12)2 = 54

Solving that equation YIELDS k = 10.

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