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1 answer:
Slope-intercept form: y = mx + b
(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)
To find the slope (m), use the slope formula:
And plug in the two points
(3, -8) = (x₁, y₁)
(6, -4) = (x₂, y₂)
(two negative signs cancel each other out and become positive)
Now that you know the slope, substitute/plug it into the equation
y = mx + b
To find b, plug in either of the points into the equation, then isolate/get the variable "b" by itself. It doesn't matter which. I will use (3, -8)
-8 = 4 + b Subtract 4 on both sides to get "b" by itself
-12 = b
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Answer:
90ft
Step-by-step explanation:
I draw it out and add a dotted line to make a triangle and rectangle then you find the area of them individually and add the two together.
Rectangle:
6*3=18
Triangle:
Add the 6+6 together to get the height of the triangle and subtract the 3 from the 15 to get the base
(12*12)
144/
72
Total area:
18+72
90ft
First, we should know exponential formula, for example exp (LnA) =A, for all A>0, and as for logarythm, LnA^p=pLnA, for all A>0, and for all value of p
so y=3^x=exp[Ln(3^x)]=exp(xLn3), let's search the graph of f(x)=exp(xLn3) f is defined in R, and its teminals limits are 0, and + infinity, f' (x) = Ln3.exp(xLn3), it is positif for all value of x, the function is strictly increasing, f(0)=Ln3, please look at the figure 1
the same method with y=2(4.5)^x, we find y=exp[Ln (2(4.5)^x)]=
exp[Ln (2) +Ln(4.5)^x)], because (LnAB=LnA + LnB), so y=2.exp[Ln(4.5)^x)]=2.exp(x[Ln(4.5)]), which domain is R, nd its teminals limits are 0, and + infinity, f' (x)=2.Ln(4.5) exp(x[Ln(4.5)]), which is positif for all value of x, the function is strictly increasing,and then f(0)=2Ln(4.5)
please look the figure 2
so y=3^x=exp[Ln(3^x)]=exp(xLn3), let's search the graph of f(x)=exp(xLn3) f is defined in R, and its teminals limits are 0, and + infinity, f' (x) = Ln3.exp(xLn3), it is positif for all value of x, the function is strictly increasing, f(0)=Ln3, please look at the figure 1
the same method with y=2(4.5)^x, we find y=exp[Ln (2(4.5)^x)]=
exp[Ln (2) +Ln(4.5)^x)], because (LnAB=LnA + LnB), so y=2.exp[Ln(4.5)^x)]=2.exp(x[Ln(4.5)]), which domain is R, nd its teminals limits are 0, and + infinity, f' (x)=2.Ln(4.5) exp(x[Ln(4.5)]), which is positif for all value of x, the function is strictly increasing,and then f(0)=2Ln(4.5)
please look the figure 2