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What is the greater of two consecutive negative integers whose product is 132?
Let x = the lesser integer and let x + 1 = the greater integer. Because product is a key word for multiplication, the equation is x(x + 1) = 132. Multiply by using the distributive property on the left side of the equation: x2 + x = 132. Put the equation in standard form and set it equal to zero: x2 + x - 132 = 0. Factor the trinomial: (x - 11)(x + 12) = 0. Set each factor equal to zero and solve: x - 11 = 0 or x + 12 = 0; x = 11 or x = -12. Since you are seems for a negative integer and reject the x-value of 11. Thus, x = -12 and x + 1 = -11. The greater negative integer is -11.
Normal Distribution Figure 1 The normal distribution reflects the various values taken by many real life variables like the heights and weights of people or the ma
a²+b²=1 a+b
log6+log-4
3/4=x/23.
If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y. (Ans:5, -2 (Not unique) Ans: Using Euclid's algorithm, the HCF (30, 72) 72 = 30 × 2 + 12
pam bought a new bedroom suit for $2588.she me a down payment of $188 and paid the remaining amount in 24 equal monthly payments .how much did she pay for each monthly payment.
Do you subtract when you add integers.
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