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X plus 2x plus 3x - 8 equals 10?
(x + 2x + 3x) - 8 = 106x - 8 = 10Add 8 to each side:6x = 18Divide each side by 6:x = 3
What is the largest whole number in the solution set for 6x plus 7 less than 3x plus 10?
6x + 7 < 3x + 106x - 3x
An account earns 3460 per month after receiving a 6 percent raise what was the accountants monthly income before raise?
let accountants monthly income before raise in income be x then x + 6x\100 = 3460 106x/100 =3460 x = 3264.15 therefore accountants monthly income before raise was 3264.15
4x plus 106x-2?
It is: 110x-2 simplified
What 2 consecutive numbers have a sum of -105?
Algebraically with X = numbers.X + (X + 1) = - 1052X + 1 = - 1052X = - 106X = - 53===========solution set
How do you solve 6x divided by 2 equals 10?
6x/2 = 106x = 10 x 2 (when taking the 2 to the RHS it is multiplied over because of the division on the LHS)6x = 20x = 20/6 (when taking the 6 to the RHS it id divided over because of the multiplication on the LHS)x = 4================================================================================Here, let me try:6x/2 = 10Reduce the fraction on the left to lowest terms.Divide numerator and denominator by 2:3x = 10Divide each side of the equation by 3:x = 10/3x = 31/3
X plus 2x plus 3x - 8 equals 10?
(x + 2x + 3x) - 8 = 106x - 8 = 10Add 8 to each side:6x = 18Divide each side by 6:x = 3
What is the largest whole number in the solution set for 6x plus 7 less than 3x plus 10?
6x + 7 < 3x + 106x - 3x
An account earns 3460 per month after receiving a 6 percent raise what was the accountants monthly income before raise?
let accountants monthly income before raise in income be x then x + 6x\100 = 3460 106x/100 =3460 x = 3264.15 therefore accountants monthly income before raise was 3264.15
What is the max amount of consecutive numbers that can be added together to make 142?
x + (x + 1) + (x + 2) + (x + 3) = 1424x + 6 = 142x = 3434 + 35 + 36 + 37 = 1425x+106x+157x+218x+289x+3610x+4511x+5512x+6613x+7814x+9115x+10516x+12017x+136There is only one sequence of consequent numbers that their sum is 142. Four consequent numbers.
What is the nth term of 104 93 82 71?
The nth term is given by: t{n} = (-3x⁴ + 30x³ - 105x² + 106x + 388)/4 for n = 1, 2, 3, 4. Alternatively, your teacher may be expecting a much simpler (also valid) solution: Each term is obtained by subtracting 11 from the previous term: t{n+1} = t{n} - 11 Which means that t0 = t1 + 11 = 104 + 11 = 115 → t{n} = 115 - 11x for n = 1, 2, 3, 4 That formula is NOT valid for any other value of n The first formula gives t1 = 104, t2 = 93, t3 = 82, t4 = 71 and gives t5 = 42, continuing with -41, -232, -603, -1244, -2263,... The second formula gives t1 = 104, t2 = 93, t3 = 82, t4 = 71 (the same first 4 terms) but gives t5 = 60 and continues with 49, 38, 27, 16, 5, ...
