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3 years ago

Rearrange each of the following in slope/y-intercept form, y = mx + b.

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

Answer:

${ \tt{3x - 6y + 8 = 0}} \\ { \tt{ 6y = 3x - 8}} \\ { \boxed{ \bf{y = \frac{1}{2} x - \frac{4}{3} }}}$

${ \tt{ - 2x - 5y - 2 = 0}} \\ { \tt{5y = - 2x + 2}} \\ { \boxed{ \bf{y = - \frac{ 2}{5} x + \frac{2}{5} }}}$

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Find the value of the expression (3⁢x-12)+12⁢(x⁢y-10) for x=4 and y=6 .

r-ruslan [8.4K]

3(4) + 12 + 12[4(6) - 10]
12 + 12 + 12(24 - 10)
12 + 12 + 288 - 120
192

5 0

3 years ago

An automobile travels past the farmhouse at a speed of v = 45 km/h. How fast is the distance between the automobile and the farm

yulyashka [42]

Answer:

$\frac{ds}{dt} = 39.586 km/h$

Step-by-step explanation:

let distance between farmhouse and road is 2 km

From diagram given

p is the distance between road and past the intersection of highway

By using Pythagoras theorem

$s^2 = 2^2 +p^2$

differentiate wrt t

we get

$\frac{d}{dt} s^2 = \frac{d}{dt} (4 + p^2)$

$2s \frac{ds}{dt} =2p \frac{dp}{dt}$

$\frac{ds}{dt} = \frac{p}{s}\frac{dp}{dt}$

$\frac{ds}{dt} = \frac{p}{\sqrt{p^2 +4}} \frac{dp}{dt}$

putting p = 3.7 km

$\frac{ds}{dt} = \frac{3.7}{\sqrt{3.7^2 +4}} 45$

$\frac{ds}{dt} = 39.586 km/h$

3 0

3 years ago

ICE Princess25 [194]

Answer:

5 x 8 /a*7

Step-by-step explanation:

8 0

3 years ago

The temperature at a mountain base camp was −3 degrees Celsius on Monday. Tuesday morning, the temperature was 2 degrees Celsius

kotegsom [21]

It is -3 to start off with, on Monday. On Tuesday morning it would be -3 - 2 degrees, = -5 degrees. By Tuesday evening it was 4 degrees lower than in the morning, so -5 - 4 degrees = -9 degrees on Tuesday evening.

Hope this helps xox :)

3 0

3 years ago

Read 2 more answers

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

3 0

3 years ago