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laila [671]
2 years ago
13

Gavin drives at a speed of 70 miles per hour. How long will it take to drive 245 miles?

Mathematics
2 answers:
katen-ka-za [31]2 years ago
3 0

Answer:

he will spend all his money on gas and not make it down the block

Margarita [4]2 years ago
3 0
3.5 Hours
Step-by-step explanation:
245÷70= 3.5

Distance
——
Time
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3 years ago
Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

A = \frac{L}{k}  - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{0}\\\frac{L}{k}  = \frac{C"}{k} \\C" = L

A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

4 0
3 years ago
PLEASE NEEED HEELPP
krok68 [10]
There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area. 
only A and E satisfy both inequality, in the overlapping shaded area.

 
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true. 
 for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities. 
Do the same for point E (5,-4)

Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area. 
8 0
2 years ago
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