Ask question

Ask question

All categories

2 years ago

8

Andy runs at a constant speed of 548.7 feet in 7.6 seconds. How far does he run in one second. Round your answer to the nearest

tenth of a second. Fill in the blank. Andy ran _____ feet in one second. *

1 answer:

IRISSAK [1]2 years ago

5 0

Answer:

72.2 feet

Step-by-step explanation:

548.7 ÷ 7.6= 72.2

You might be interested in

I really need help and this not working

Nostrana [21]

Answer:

Help with what

Step-by-step explanation:

6 0

2 years ago

A bag containing food has mass 3.813 kg. When the bag is empty its mass is 257 g. Find in kilograms the mass of the fruit

Elden [556K]

Answer:

I think it is 0.257 kg

Step-by-step explanation:

To change units from the gram to kilogram, you need to divide by 1000, because 1kg=1000g:

257÷1000= 0.257 Kg

5 0

2 years ago

A line includes the points (0,0) and (1,5). What is its equation in slope-intercept form

Kipish [7]

9514 1404 393

Answer:

y = 5x

Step-by-step explanation:

The point (0, 0) tells you the line goes through the origin, so represents a proportional relationship. The slope is simply the ratio of y to x, or 5/1 = 5.

The equation of the line is ...

y = mx + b . . . . . . line with slope m and y-intercept b

y = 5x . . . . . . . . . . line with slope 5 and y-intercept of 0 (the origin)

5 0

2 years ago

The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$

When t = 0, N = 400

$400 = 725 - C e^{-k*0}\\400 = 725 - C\\C = 725 - 400\\C = 325$

When t = 3, N = 650

$650 = 725 - (325 * e^{-3k})\\325 * e^{-3k} = 75\\e^{-3k} = 75/325\\e^{-3k} = 0.23\\-3k = ln 0.23\\-3k = -1.47\\k = 1.47/3\\k = 0.49$

The equation for the population becomes:

$N = 725 - 325 e^{-0.49t}$

At t = 5, the population becomes:

$N = 725 - 325 e^{-0.49*5}\\N = 725 - 325 e^{-2.45}\\N = 696.95\\N(5) = 697$

N(5) = 700 ( to the nearest 50)

6 0

3 years ago

NEED ANSWERS PLEASE!

Harman [31]

I’m pretty sure it’s the second answer? but not for sure i’ve done this before though

5 0

3 years ago