Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
Answer:
I believe x = 34
Step-by-step explanation:
3 × 34 = 102
102 - 6 = 96
2 × 34 = 68
96 + 68 + 88 + 108 = 360
Answer:
Step-by-step explanation:
The given equation is 
We substitute the ordered pairs to see which ones satisfies the equation.
For (-1,-8), we put x=-1 and y=-8 to get:
This is not true.
For 
we put x=-1 and 
to get
This is true
For (3,3/2) we have 2(3)+6(3/2)=15
This implies 6+9=15
15=15
This is also True
For (-5,25/6), we have 2(-5)+6(25/6)=-10+25=15
This is also true.
Answer:
I am pretty sure your answer is right
Step-by-step explanation:
Answer:
b I did the test already
Step-by-step explanation:
B
