Answer:
a) T(t)= 100e^(-0.0513t)
b) See attachment for graph
y-intercept is inital temperature
domain is t≥0, range is 0 <T≤100
c) 44.9min
Step-by-step explanation:
a) at t=0, T=100
at t=1, T=95
T(t)= Ae^-kt
100=Ae^(-k×0)
A=100
95=100e^(-k
k=0.0513
T(t)= 100e^(-0.0513t)
b) as time reaches infinity, temperature of pudding becomes approximately equal to 0. x-axis is the asyptote of the graph
c) 10= 100e^(-0.0513t)
t= 44.9 min
Answer:
The answer is "GK=37, GH=14, and JK=19 ".
Step-by-step explanation:
In the question, an attachment file is missing.so, please find its attached file.
Given values:
GJ=4x+2
HK=6x-1
GK =37
Find:
HJ=?
GK = GJ+JK
= GJ+(HK-HJ)
= 4x+2+((6x-1)-x)
= 4x+2+6x-1-x
=
9x+1...(a)
Solve equation (a) and put the value of x in the equation:
=9(4)+1
=36+1
=37
find GH and JK=?
Given:
GK=37
Formula:
GH = GJ-HJ
= 4x+2- x
= 3x+2....(b)
Solve equation (a):
GK= 9x+1
37=9x+1
36=9x
x=4....(c)
put the value of equation (b) in equation (c)
GH = 3x+2
= 3(4)+2
= 12+2
= 14
Formula:
JK= HK-HJ
= 6x-1- x
= 5x-1
put the value of x in the above equation:
= 5(4)-1
= 20-1
=19
Answer: 137
Step-by-step explanation:
The given expression:
We need to find the value of the expression when x=4.
So we just use the substitution property and substitute the value of x= 4 in the given expression , we get
Hence, the value of the expression at x=4 is 137.
Answer:
Marla didn't use the right steps to complete the square. Maria made a mistake in step 1, she put 8x instead of -2x
Step-by-step explanation:
we have
This is a vertical parabola open downward
The vertex is a maximum
Find the vertex
step 1
Factor the leading coefficient -4
step 2
Complete the square
step 3
step 4
Rewrite as perfect squares
the vertex is the point (1,-21)
so
The maximum value of the quadratic equation is (1,-21)
therefore
Marla didn't use the right steps to complete the square. Maria made a mistake in step 1, she put 8x instead of -2x