I must assume that you meant t=1 (not t=?1). If t=1, here's what we'd do:
1. Find the x and y values corresponding to t=1. They are:
x=(1)^7+1=2 and y=(1)^8+1=2. (Please note: write t^8 instead of t8, and write t^7 instead of t7.)
2. The slope of the tangent line to the graph is
dy/dt 8t^7+1
dy/dx = ---------- = ---------------- with 1 substituted for t
dx/dt 7t^6
Thus, dy/dx (at t=1) = 9/7
3. Now we have both a point (2,2) on the graph and the slope of the tangent line to the curve at that point: 9/7
4. The tangent line to the curve at (2,2) is found by using the point-slope formula:
y-y1 = m(x-x1)
which comes out to y-2 = 9/7(x-2), or 7y-14 = 9(x-2). You could, if you wished, simplify this result further (e. g., by solving for y in terms of x).
Answer:
90 days.
Step-by-step explanation:
dy/dt = - 0.0032y
dy = -0.0032y dt
dy/y = -0.0032 dt
Integrating:
ln y = -0.0032t + c where c is some constant
y = e^(-0032t) + e^c
y = A e^(-0032t) where A is a constant.
At time 0 we can let y = 1 so in that case A = 1
When there is 0.75 of y left:
0.75 = e^-0032t
ln 0,75 = -0032 t
t = ln 0.75 / -0032
t = 90.
Answer:
45
Step-by-step explanation:
because the mass
Answer:
Step-by-step explanation:
y = 3x - 5.....input values are x and output values are y
so if ur input (x) is 2.....sub 2 in for x and find ur output (y)
y = 3x - 5
y = 3(2) - 5
y = 6 - 5
y = 1 <=== ur output value
Answer:
y = (5/7)x + 3
Step-by-step explanation:
We use the form y = mx + b. Substituting (5/7) for m, 7 for x and 8 for y, we get:
8 = (5/7)(7) + b, or 8 = 5 + b. Thus, b = 3, and the desired equation is
y = (5/7)x + 3
