Answer:
y=-3x+10
Step-by-step explanation:
Use point-slope form first.
y-y1=m(x-x1)
Plug the numbers into that.
y+2=-3(x-4)
Distribute and simplify to convert it to y=mx+b
y+2=-3x+12
y=-3x+10
Answer:
A) 0.106
Step-by-step explanation:
8e^3x + 4 = 15
Subtract 4 from each side
8e^3x + 4-4 = 15-4
8e^3x = 11
Divide each side by 8
8/8e^3x = 11/8
e^(3x) = 11/8
Take the natural log of each side
ln(e^(3x)) = ln(11/8)
3x = ln(11/8)
Divide by 3
3x/3 = 1/3 ln(11/8)
x = 1/3 ln(11/8)
x =.106151244
To the nearest thousandth
x = .106
Answer: 36th customer.
Step-by-step explanation: 36 is divisible by both 9 and 12, and is the least common multiple.
Answer:
0.05 + 0.1y = 0.12 + 0.06y
Step-by-step explanation:
Here, we want to find y which is the heat in which the amount of mercury in each of the water bodies is same.
Now for the first water body:
Initial is 0.05 ppb and after y years, we have a rise of 0.1 * y = 0.1y ppb
So the amount of mercury in ppb after y years would be; 0.05 + 0.1y
For the second water body;
Initial is 0.12 and a rise of 0.06 ppb per year for y years. The rise would be 0.06 * y = 0.06y
So total amount of mercury here is 0.12 + 0.06y
So to find y which is the year the amount of mercury in each water body is same, we simple equate both and that would be;
0.05 + 0.1y = 0.12 + 0.06y
Answer:
The first graph. (an upside down U)
Step-by-step explanation:
The leading coefficient of a polynomial determines the direction of the graph's end behavior.
A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.
A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.
This graph has two evens. Because its negative, only one is possible - the first graph.
The other two graphs are odd with both starting down on the left and point up on the right which is a positive leading coefficient. These are not possible graphs.
The first graph is the solution.
