1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vlad1618 [11]
3 years ago
12

2a - 5d = 30 dove for d

Mathematics
2 answers:
Slav-nsk [51]3 years ago
6 0
I hope this helps you




2a-30= 5d


d=2a-30/5


d=2a/5-6
AfilCa [17]3 years ago
5 0

2a - 5d = 30 solve for d

subtract 2a to both sides

2a - 5d - 2a = 30 - 2a

simplify

-5d = 30 - 2a

divide both sides by -5

-5d/-5 = (30 - 2a) / -5

simplify

d = -6 +0.4a

d = 0.4a - 6

You might be interested in
The last question for my test help
alisha [4.7K]

3 2/6

5/6*4=20/6

20/6= 3 2/6 OR 3 1/3

8 0
3 years ago
Find the surface area of the prism
Bess [88]

Answer:

it would be 124mm

Step-by-step explanation:

you multiply the base x height

3 0
3 years ago
Help please quick! i need to find all the missing angles and side
Molodets [167]

Answer:

I think the answer is 73.

Step-by-step explanation:

did it state the volume?

wait its a right angle, so its 90 degrees?

8 0
2 years ago
C. Find the value of x below:<br> 15.r + 5<br> -2 + 16.
nydimaria [60]

Answer:

7

Step-by-step explanation:

(15x + 5) and (-2 + 16x) represents the measures of corresponding angles.

Measures of corresponding angles are equal.

Therefore,

- 2 + 16x = 15x + 5

16x - 15x = 5 + 2

x = 7

5 0
3 years ago
Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua
goblinko [34]
<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

D(t)=\frac{5t}{1+3t}

So, the integral will be:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dx

So, integrating we have:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx

Performing a change of variable, we have:

1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}

Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]

Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0
3 years ago
Other questions:
  • Are BD and CE congruent? Explain.
    12·1 answer
  • 15 points please help
    5·1 answer
  • ALGEBRA HELP !!!!! PHOTO ATTACHED!!!!! Function f(x), represented by the graph, is _____. The value of f(150), rounded to the ne
    8·2 answers
  • What is arc BC equal to?
    11·1 answer
  • I need to go 1,373 miles but the plane travels at 575 miles per hour. how long would it take me to get to my destination?
    9·1 answer
  • What is the quotient of 4536 and 36?
    5·2 answers
  • What is the term for this question?
    12·1 answer
  • Find the sides marked with letters:<br>m = ? z = ?<br><br>​
    9·1 answer
  • Who ever gets this right will get a brainlest and a good rating
    9·2 answers
  • A car can travel 120 miles on 4 gallons of gasoline. How much gasoline will it need to go 510 miles?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!