To find 4 times the sum of y and 7, you first calculate the sum of y and 7, which is y + 7. Then, you multiply this sum by 4 to get the final result. Therefore, the expression for 4 times the sum of y and 7 is 4(y + 7), which simplifies to 4y + 28.
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Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25
17 goes in 119 7 times.
2 *5y
Call your number 10x + y. x = y + 2 and 10x + y = 4 + 6(x + y) Substitute y + 2 for x: 10(y + 2) + y = 4 + 6((y + 2) + y) This simplifies to 10y + 20 + y = 4 + 6y + 12 + 6y, ie 20 - 16 = 12y - 11y so y = 4 and x = 6 Your number is 64, which is indeed 4 more than the sum of its digits.
y=mx+b y = -(7/4)x + 7