All categories

4 years ago

How would I solve -6.11 + b = 14.321

Mathematics

2 answers:

Studentka2010 [4]4 years ago

8 0

Answer:

b=20.431

Step-by-step explanation:

you just add 6.11 to both sides. on the left side the -6.11 and +6.11 cancel out and on the right side you will just add 6.11 to 14.321. when you add 14.321 and 6.11 you get 20.431. therefore, b=20.431

klasskru [66]4 years ago

4 0

You would do the inverse of -6.11 to cancel it out. So would be adding 6.11 to 14.321. That would give you b=20.431.

You might be interested in

Find the sum of 5x-y and (2xy+xy)

devlian [24]

Since neither 5x nor -y has a "like term" in 2xy+xy, the sum of these quantities is simply

2xy+xy + 5x - y = 3xy + 5x - y

You might want to write this as y(3x -1) + 5x.

8 0

3 years ago

Show how to solve 6/15=2/?

Bas_tet [7]

There are two way sof doing this question:
1st solution is : 
 6/15 = 2/5
because 6/15 is divisible by 3 and it is done by this form :
 3 X 2 = 6
 and
 3 X 5 = 15
2nd solution is:
6/15 = 2/?
here we suppose that the missing value is x:
 6/15 = 2/x
you shift the denominators and then multiply them by the their numerators:
6 * x = 15 * 2
 6x = 30
now you divide 6 on both sides:
 6x/6 = 30/6
 x = 5

6 0

3 years ago

Read 2 more answers

Help, offering BRAINLIEST AND POINTS. One year ago, Lindsey deposited $250 into a savings account. Her balance is now $253. Two

andrew11 [14]

Answer:

$257.50

Step-by-step explanation:

Because it has a higher amount.

8 0

3 years ago

Using complete sentences, describe the net of a rectangular prism with a length of 12 centimeters, a width of 9 centimeters, and

LUCKY_DIMON [66]

Answer:

it would be 12 multiple by 9 then add the 5 centimeters

Step-by-step explana

8 0

4 years ago

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres

topjm [15]

Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$ . As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$ . This means that at time $t=0$ we have $U(0)=u$ , so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$ ,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$ :

$0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}$

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0

3 years ago