If the sum of an integer and 7 more than the next consecutive integer is 66 then the integers are 29 and 30.
Given that the sum of an integer and 7 more than the next consecutive integer is 66.
Integer is a number that is written without a fraction component. It can be positive as well as negative.
let the first integer be x.
Consecutive integer will be x+1.
Sum of integer and 7 more than the next consecutive integer is 66.
Sum according to the given information=x+x+1+7
=2x+8
2x+8=66
2x=66-8
2x=58
x=58/2
x=29
Next integer=29+1-30.
Hence if the sum of an integer and 7 more than the next consecutive integer is 66 then the integers are 29 and 30.
Learn more about integers at brainly.com/question/17695139
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When you are checking your answers, your formula should read out as 9(2.222) - 8 = 12, which is simply filing in the x value that you calculated (it comes out to 11.998 which is correct). You wrote it out as 9(2.22) - 8 = -6, which is why your checking process was incorrect
Answer:
5 1/3 or 5.33333333333333333333333333333333333333333333333333333
Step-by-step explanation:
to find the scale factor, you must divide the two equal sides.
which equals
or 0.75
Now, the way to get the value of x is to divide 4 by 3/4 0r 0.75
Which equals
5.333 repeating or 5 1/3
Hope this helps. If you have any follow-up questions, feel free to ask.
Have a great day! :)
Answer:
what is the rest of the equation?
Step-by-step explanation:
Answer:
Step-by-step explanation:
26% is (26-20)/(50-20) = 6/30 = 1/5 of the way between 20% and 50%. That means 1/5 of the solution is 50% acid.
50% acid: 1/5 · 100 mL = 20 mL
20% acid: 100 mL -20 mL = 80 mL
Delbert must mix 80 mL of 20% acid and 20 mL of 50% acid.
_____
Maybe you'd like to see an equation. Let x represent the amount of 50% acid required. Then 100-x is the amount of 20% acid needed. The amount of acid in the mix is ...
0.50(x) +0.20(100 -x) = 0.26(100)
(0.50 -0.20)x = (0.26 -0.20)100 . . . . subtract 0.20(100)
x = (0.26 -0.20)/(0.50 -0.20)×100 = 20
This last expression should look a lot like the one we started with in this answer. It shows you how you can almost write down the answer to mixture problems without a lot of work.
