◆ Define the variables:
Let the calorie content of Candy A = a
and the calorie content of Candy B = b
◆ Form the equations:
One bar of candy A and two bars of candy B have 774 calories. Thus:
a + 2b = 774
Two bars of candy A and one bar of candy B contains 786 calories
2a + b = 786
◆ Solve the equations:
From first equation,
a + 2b = 774
=> a = 774 - 2b
Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie
◆ Find caloric content:
Caloric content of candy B = 254 calorie
Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
Answer is 67.5
To get this you must first turn 54% into a decimal.
54%---> .54
125*.54= 67.5
So 54% of 125 is 67.5 <span />
Hello,
Let me try out the solution for you.
Consider the below scenarios for the equation y = |x+5|-|x-5|
Case 1:
when x more than or equal to 5
then y=(x+5)-(x-5) = 10
hence y=10
Case 2:
when -5<x<5
y=(x+5)-(-(x-5)) = 2x
y=2x so y can take 9 values corresponding to x={-4,-3,-2,-1,0,1,2,3,4}
Case 3:
when x less than or equal to -5
y= -(x+5)-(-(x-5))
y=-10
Hence if we combine all 3 cases we get that y can take total of 11 values.
Answer:
-3/2,collinear.
Step-by-step explanation:
slope of PQ=(0-3)/(2-0)=-3/2
eq. of line PQ is
y-3=-3/2(x-0)
y-3=-3/2 x
or2y-6=-3x
or 3x+2y=6
if R(4,-3) lies on it,then
3(4)+2(-3)=6
12-6=6
or 6=6
which is true.
Hence points P,Q,R are collinear.
and line has aslope=-3/2
