◆ 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
I’m not sure I only answer these for the points :)
Domain is the x values you can use
range is the y values
a.
the domain
hmm, seems to be all real numbers except for at x=0, it gets really close tho
so D=(-∞,0)U(0,∞)
range is all real numbers except for y=0, it gets really close tho
R=(-∞,0)U(0,∞)
b.
domain
that empty circle means something like < or >
goes from 5 to 6, not including 5
so domain is D=(5,6]
range is from -4 to 2, not including -4
range is R=(-4,2]
The answer would be A. m= -3/5 and b=-4/5
"m= -5-1 divided by 7--3
or, -6/10
or, -3/5.
to find the (B)
<span><span>(-3,1). y=mx+b or 1=-3/5 × -3+b, or solving for b: b=1-(-3/5)(-3). b=-4/5.</span><span>(7,-5). y=mx+b or -5=-3/5 × 7+b, or solving for b: b=-5-(-3/5)(7). b=-4/5.</span></span>See! In both cases we got the same value for b. And this completes our problem.
<span><span>The equation of the line that passes through the points(-3,1) and (7,-5)is</span><span>y=-3/5x-4/5"</span>
Source: webmath</span>