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3 years ago

3 loq, 6) - 2 log (5x) = 2

Mathematics

1 answer:

Scrat [10]3 years ago

4 0

Answer:

is this the answer im sorry i did not put steps i was in a hurry to go to the next class.

Step-by-step explanation:

y=10(log(5x)+1)/9lq(y,x)

l=0

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2x+4y=6<br> 3x=12-6y<br> Solve by elimination

faltersainse [42]

Answer:

no solution

Step-by-step explanation:

2x + 4y = 6......reduces to x + 2y = 3

3x = 12 - 6y....reduces to x = 4 - 2y...rearranged is x + 2y = 4

so now we have :

x + 2y = 3

x + 2y = 4

ok....I dont have to go any farther to know that this has no solution because ur equations have the same slope and different y int, this means ur lines are parallel and have no solution because they never cross each others path.

5 0

3 years ago

Read 2 more answers

0.2y=0.5x+0.1 please solve with math thank you

Cerrena [4.2K]

<span>"Simplifying 0.2y = 0.5x + 0.1
Reorder the terms: 0.2y = 0.1 + 0.5x Solving 0.2y = 0.1 + 0.5x Solving for variable 'y'.
Move all terms containing y to the left, all other terms to the right.
Divide each side by '0.2'. y = 0.5 + 2.5x Simplifying y = 0.5 + 2.5x"</span>

7 0

3 years ago

Cindy wants to predict how much energy she will use to heat her home based on how cold it is outside. The table below shows the

Mademuasel [1]

Cindy wants to predict how much energy she will use to heat her home based on how cold it is outside. The table below shows the mean amount of gas per day (in cubic meters) that Cindy used each month and the average temperature that month (in degrees Celsius) for one heating season Well

B. is The Answer :)

4 0

2 years ago

Read 2 more answers

Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

hence the complete solution

y = C.F. + P.I.

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

7 0

2 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

earnstyle [38]

Answer:

(i). Average speed = 75 km per hour

(ii). Average speed = 75 km per hour

Step-by-step explanation:

This question is incomplete; here is the complete question.

A train traveled from station A to station B at an average speed of 80 kilometers per hour and then from station B to station C at an average of 60 kilometers per hour. If the train did not stop at station b, what was the average speed at which the train traveled from station A to C?

(i). The distance that the train traveled from station A to station B was 4 times the distance that train traveled from station B to station C.

(ii). The amount of time it took to the train to travel from station A to station B is 3 times the amount of time that it took the train to travel from station B to station C.

(i) Let the distance between station B and station C = x km

So the distance between Station A and station B = 4x km

Therefore, time to travel the distance x km with 80 km per hour = $\frac{\text{Distance traveled}}{\text{Speed}}$