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Georgia [21]
3 years ago
6

a red car is traveling 15 miles per hour faster than a blue car. if the red car travels 180 miles in the time needed for the blu

e car to travel 144 miles, find speed of each car
Mathematics
1 answer:
vlabodo [156]3 years ago
5 0
Their distance differs by 15 miles for each hour of travel time. Thus, the travel time of interest is
.. (180 -144) miles/(15 miles/hour) = 2.4 hour

The red car's speed is 180 mi/(2.4 h) = 75 mi/h.
The blue car's speed is 144 mi/(2.4 h) = 60 mi/h.
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enot [183]
A straight line needs two pieces of information to be identified, a gradient and a y-intercept (technically any point will do but the y-intercept is particularly convenient if we have it).

The gradient is calculated by taking two points on the line, and dividing the change in y-coordinate by the change in x-coordinate between them. I'm going to take the points (0,-3) and (2,-2).

The change in y-coordinate is (-2) - (-3) = 1

The change in x-coordinate is (2) - (0) = 2.

Gradient = m = 1/2

Next we identify the y-intercept, the value of y when x = 0. This value is -3, and we call it c.

The equation of a line in slope-intercept form is y = mx + c. Slotting in the values for m and c we have ascertained, we find that the equation of this line is:

y = (1/2)x - 3

I hope this helps :)
7 0
3 years ago
Resolver las siguientes desigualdades<br> 3y&lt;6<br> -x/2&gt;2<br> Y+8&gt;9<br> -3x &lt;15
Crazy boy [7]

Answer:ndbdnd ndbdnd 192

Step-by-step explanation:

BBB as haha

8 0
4 years ago
A bacteria population is 3000 at time t = 0 and its rate of growth is 1000 · 8t bacteria per hour after t hours.
julia-pushkina [17]
P=3000x e^( 1000 x 8t )is the function
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4 0
3 years ago
During the period from 1790 to 1930, the US population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this
alex41 [277]

Answer:

Step-by-step explanation:

Given that during  the period from 1790 to 1930, the US population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this period, P(t) remained close to the solution of the initial value problem.

\frac{dP}{dt} =0.03135P =0.0001489P^2, P(0) = 3.9

a) 1930 population is the population at time t = 40 years taking base year as 40

We can solve the differential equation using separation of variables

\frac{dP}{0.03135P – 0.0001489P^2 } =dt\\\frac{dP}{-P(0.03135 – 0.0001489P } =dt

Resolve into partial fractions

\frac{31.8979}{P} -\frac{0.00474}{0.0001489P-0.03135}

Integrate to get

ln P -0.00474/0.0001489  (ln (0.0001489P-0.03135) = t+C

ln P -31.833  (ln (0.0001489P-0.03135) =t+C

\frac{P}{( ( (0.0001489P-0.03135)^{31.833}  } =Ae^t

Limiting population would be infinity.

6 0
4 years ago
A set of 125 golf scores are normally distributed
Phantasy [73]

Answer:

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3 0
3 years ago
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