The sum of a number and 19 is at least 8 using inequalities

Answers

Answer 1
Answer: n+19 >= 8
since their is no greater than or equal sign i did that^ hope u understand
Answer 2
Answer: n+19>=8. :)Have fun with your homework ;)

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What factors do 28 and 44 have in common

Answers

The factors of 28 : 1, 2, 4, 7, 14, 28The factors of 44 : 1, 2, 4, 11, 22, 44so the common factors of 28 and 44 are : 1, 2, 4

What is 20/3 equal to?

Answers

20/3 is equal to ------?


First we will divide numerator by the denominator.

20 ÷ 3 = 6.66666666667

Ok, Now we will see how many times 3 goes in to 20. 

3 × 6 = 18.
 
There will be 2 left. SO we will set our fraction like a mixed number:-



6 2/3


Hope I helped:P



I recently took a test in my Algebra 1 class. This was on the test (that I most likely failed)and it said something about vertical motion. The problem is below. Try to solve it, please! Thanks!2=-t^2+4+5

Answers


OK.                                         2 = -t² + 4t + 5

Subtract 2 from each side:     0 = -t² + 4t + 3

Multiply the right side by  -1 :  0 = t² - 4t - 3

Unfortunately, this can't be factored.  The only other way I know
to find the solutions to a quadratic equation is with the use of the
quadratic formula.  When you do that, you get these solutions:

     t = 2 + √7 = about  4.65
and
     t = 2 - √7 = about  -0.65 .

If this equation comes from a problem that involves real people throwing
real things into the air and making vertical motion, then you want only the
positive solution.


The combined age of three relatives is 120 years. James is three times the age of Dan, and Paul is two times the sum of James and Dan. How old is each relative

Answers

The ages of each relative are 30, 10 and 80 years respectively.

  • Let James' age be J.
  • Let Dan's age be D.
  • Let Paul's age be P.

Translating the word problem into an algebraic expression, we have;

J + D + P = 120 ......equation 1

J = 3D          .......equation 2

P = 2(J  + D)   ......equation 3

Substituting eqn 2 and 3 into eqn 1, we have;

3D + D + 2J + 2D = 120\n\n6D + 2(3D) = 120\n\n6D + 6D = 120\n\n12D = 120\n\nD = (120)/(12)

Dan, D = 10 years

For James;

J = 3(10)

James, J = 30 years.

For Paul;

P = 2(J  + D)

P = 2(30 + 10)\n\nP = 2(40)

Paul, P = 80 years.

Therefore, the ages of each relative are 30, 10 and 80 years respectively.

Read more: brainly.com/question/6463206

Answer: Your answer is 60

Step-by-step explanation:

How can I adjust a quotient to solve a division problem

Answers

for example you have five times twevle people so that's 60people. 5 times 12 is 60 and 12 times 5 is 60 so 60 dividedby 5 is 12 and 60 divided 12 is five. that's what division can be like

Final answer:

To adjust a quotient in a division problem, you can multiply both the dividend and divisor by the same number.

Explanation:

To adjust a quotient in order to solve a division problem, you can multiply both the dividend and divisor by the same number. This will not change the value of the quotient, but it will make the division easier to solve.

For example, if you have 12 ÷ 3 and you want to adjust the quotient, you can multiply both 12 and 3 by 10 to get 120 ÷ 30. The quotient remains the same, but now you can easily divide 120 by 30 to get the answer, which is 4.

In summary, adjusting a quotient involves multiplying both the dividend and divisor by the same number to simplify the division problem.

Learn more about Adjusting a quotient in division problems here:

brainly.com/question/32831149

#SPJ11

How many cups are there in 6.25*10^22 nL? Hint: nL is the abbreviation for nanoliter.

Answers

A nano is essentially one-billionth or 1/1,000,000,000.
Since writing all these zeroes is inefficient and can easily lead into errors, we'll just use scientific notation instead.
A nanoliter is 10^(-9) liters.

So we have 6.25 * 10^(22) *  10^(-9) = 6.25* 10^(13) liters.

A liter is 4.22675 cups. So now we can multiply 4.22675 by 6.25 * 10^13.
Use a calculator... you get 26.4171875 * 10^13.

*Note: There may be slight precision errors. This is inevitable when you're working with very large numbers, as the smallest initial change can spiral into much larger differences.