This problem is a problem using
d = r * t
First part of the trip
d = 96 miles
r = ??
t = ??

Second part of the trip
d = 15 miles
r = r - 5
t = 4 - t

Solve for r first.
96/r + 15/(r - 5) = 4 hours multiply through by r and r - 5
96(r - 5) + 15r = 4*(r - 5)(r) Remove the brackets on the left
96r - 480 + 15r = 4*r*(r - 5) Collect like terms on the left.
111r - 480 = 4r(r - 5) Remove the brackets on the right.
111r - 480 = 4r^2 - 20r Bring the left side over to the right side.
0 = 4r^2 - 20r - 111r + 480 Collect like terms.
0 = 4r^2 - 131r + 480

The only way I could do this was to use the quadratic equation.

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$

a = 4

b = - 131

c = 480

$\text{x = }\dfrac{ -(-131) \pm \sqrt{(-131)^{2} - 4*4*480 } }{2*4}$
$\text{x = }\dfrac{ 131 \pm \sqrt{(17161) - 7680 } }{8}$
$\text{x = }\dfrac{ 131 \pm \sqrt{9481 } }{8}$
$\text{x = }\dfrac{ 131 \pm 97.37 }{8}$

x = (131 + 97.38)/8
x = 28.55
x = (131 - 97.37)/8
x = 4.20

4.2 isn't going to work because the slower speed has to be larger than 5, otherwise you will get a negative speed (as this solution will give), so 4.20 is an extraneous solution.

r = 28.55 for the first part of the trip and
r = 23.55 for the second part of the trip.

Check.
Let's see if the times add up to 4
d/r = t
96/28.55 = 3.36
15/22.75 = 0.64

The times do = 4, so the rates are correct.

Answers
r1 = 28.55 for the 96 mile trip.
r2 = 23.55 for the 15 mile trip.

8 0

2 years ago

Which is longer 2 meters or 200 cent a meters or 2.05 meters

Yuri [45]

Solving this requires use of conversions

we already know that
2.05 m > 2 m

all that's left is the question
200 cm vs. 2.05 meters

I we know our equivalencies, we know that
100 cm = 1 m

if this is so then
200 cm = 2 m

we already know that
2.05 m > 2 m

so it must also be that
2.05 m > 200 cm

$2.05 \: meters$

6 0

3 years ago

H(n) = 3^n+3; Find h (3n)

olya-2409 [2.1K]

Answer:

The option which you chose is absolutely right.

Step-by-step explanation:

h(n) = 3^ n + 3

h(3n) = 3 ^3n + 3

Hope it will help :)

5 0

2 years ago