−23⋅(7+2)+11 = −23⋅ +11 = +11 =

Define a variable for the unknown quantity below.

stira [4]

Step-by-step explanation:

Gladys blister is luggage

5 0

3 years ago

I need help with #3. I would like it if you could explain how to do it because there are 2 more that are similar to #3. Thank yo

zysi [14]

Answer:

raha of my life with the exception in my case the first two of the year old one is that it will help you are not able the first of

5 0

3 years ago

Solve the system. x + y - z = 17 y +z = 1 z = -3

Anna35 [415]

Answer:

(10, 4, -3)

Step-by-step explanation:

z = -3

y + z = 1

y - 3 = 1

y = 4

x + y - z = 17

x + 4 - -3 = 17

x + 4 + 3 = 17

x + 7 = 17

x = 10

(10, 4, -3)

3 0

2 years ago

Read 2 more answers

24-4x=32 Show your work

Mandarinka [93]

Answer:

-2

Step-by-step explanation:

24 - 4x =32

-4x =32 - 24

-4x=8

x=-8 :4

x=-2

8 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