All categories

2 years ago

What is the value of b 2b + 8 − 5b + 3 = −13 + 8b − 5

Mathematics

2 answers:

eimsori [14]2 years ago

8 0

Step-by-step explanation:

2b-5b+8+3 = - 13 - 5 +5b

-3b + 11 = -18 +5b

11+18 = 5b + 3b

29 = 8b

29/8 = b

3.625 = b

Yuri [45]2 years ago

3 0

Answer:

See below.

Step-by-step explanation:

2b + 8 − 5b + 3 = −13 + 8b − 5

Reorder like terms.

2b-5b+8+3=8b-13-5

Combine those like terms. Then solve.

-3b+11=8b-18

-11 -11

-3b=8b-29

-8b -8b

-11b=-29

/-11 /-11

b=2.64

-hope it helps

You might be interested in

What is 66 over 108 in the lowest term

ella [17]

answer

lowest form of 66\108

=11\18

5 0

2 years ago

Read 2 more answers

Select the correct answer. Consider this absolute value function. f(x)= |x-5| How can function f be written as a piecewise funct

faltersainse [42]

Option (c) is correct according to me I think it’s helpful

6 0

2 years ago

Read 2 more answers

A tabletop has an area of 12 square feet and a perimeter of 16 ft. What are the dimensions of the table? 2 ft by 6 ft 2 ft by 8

ICE Princess25 [194]

Answer:2ft by 6ft

Step-by-step explanation:

4 0

3 years ago

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

6 0

2 years ago

What is the decimal form of 8 and 39/40

steposvetlana [31]

8.975 is the correct answer

6 0

2 years ago

Read 2 more answers