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Diano4ka-milaya [45]

4 years ago

14

Two race cars,car x an y,are at the starting point of a two km track at the same time.car x and car y make one lap every 60 s an

d every 80 sec respectively.
How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?

1 answer:

tino4ka555 [31]4 years ago

4 0

Answer is attached.

Please check the image.

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Mod 8 Honors Assignment - Sequences and Series

Gekata [30.6K]

The numbers in the sequence will be 40, 80, and 160.

<h3>How to get the values?</h3>

Let x be the first number in the sequence, so the first three numbers are {x, 0.5x, 0.5²x}.

Then

{x/4 + 10, 0.5x - 10, 2(0.5²x) + 10}

0.5x - 10 = x/4 + 10 + c = x/2 - 10 = x/4 + 10 + c

2(0.5²x) + 10 = 0.5x - 10 + c = x/2 + 10 = x/2 - 10 + c

Solve the second equation for c :

x/2 + 10 = x/2 - 10 + c

c = 20

Substitute this into the first equation and solve for x :

x/2 - 10 = x/4 + 10 + 20

x/4 = 40

x = 160

Then the terms are {160, 80, 40}

Learn more about sequence on:

brainly.com/question/6561461

#SPJ1

4 0

2 years ago

Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production of these calcul

kondaur [170]

Answer:

Calculators from the beginning of the third week to the end of the fourth week = 4048.

Step-by-step explanation:

We know that the rate of production of these calculators after t weeks is given by

$\frac{dx}{dt} =5000(1-\frac{100}{(t+10)^{2}})$

To find the number of calculators that have been produced in a period, we need to take the integral of the function above; the desired time is t=2 (beginning of third week) to t=4 (end of the fourth week). Therefore, the number of calculators produced in the given time is

$\int\limits^4_2 {\frac{dx}{dt} } \, dt = \int\limits^4_2 {5000(1-\frac{100}{(t+10)^{2} }) } \, dt$

Substitute $t+10=u$ and $dt=du$ , observe that the limits of integration will change

$\int\limits^4_2 {\frac{dx}{dt} } \, dt => \int\limits^{14}_{12} {\frac{du}{dt} } \, dt$

$5000\int\limits^{14}_{12} { 1-\frac{100}{u^{2} } } \, du$

$5000(u+100u^{-1})\left \{ {{14} \atop {12}}\right.\\5000(2+\frac{100}{14}-\frac{100}{12} )\\4047.62$ ≈ 4048

4 0

3 years ago

What is the sum of 12 – 5i and –3 + 4i?

8090 [49]

12 - 3 =9 and -5i+4i=1i or i so it is {9-i}

6 0

3 years ago

Read 2 more answers

Help me with this question pleaseeee!!!

Butoxors [25]

We have that
the measure of angle ∅=7*pi/6
the measure of its reference angle is?
remember that pi=180°
then
7*pi/6=7*180/6=210°------------> III Quadrant
Part A) its reference angle is 210°-180°=30°
Part B) sin(30°)=-0.5--------> negative because belongs III quadrant

5 0

3 years ago

[(3 x 2) - 2 x 1.25]

kondor19780726 [428]

Answer:

3.5

Step-by-step explanation:

[(3 x 2) - 2 x 1.25] = [6 - 2 x 1.25]

= [6 - 2.5]

= 3.5

5 0

3 years ago

Read 2 more answers