Answer:
so the answer will be 1
Step-by-step explanation:
An oil tanker can be emptied by the main pump in 5 hours.
An auxilary pump can empty the tanker in 14 hours.
If the main pump is started at 7pm, when should the auxilary pump be started
so that the tanker is emptied by 11pm?
:
Let t = no. of hrs to run the Aux pump
:
The main pump will run 4 hr (7 - 11 pm)
:
Let the completed job = 1; (an empty Tanker)
:
The shared work equation
4%2F5 + t%2F14 = 1
Multiply equation by 70 to get rid of the equation, results:
14(4) + 5t = 70
56 + 5t = 70
5t = 70 - 56
5t = 14
t = 14%2F5
t = 2.8 hrs to run the aux pump
:
2.8 hr = 2 + .8(60) = 2 hrs 48 min
:
Subtract 2:48 from 11:00 = 8:12 PM start the aux pump
;
;
Check solution
4/5 + 2.8/14 =
.8 + .2 = 1
80% of 20 is 80/100 x 20 = 16
So, kelly has used 16 stamps, and has 4 left
Answer:
10x + 16
Step-by-step explanation:
Doing the distributive property, we get the answer 10x + 16.
<span>I can be wrong, but I think that the answer is: To find the values of p, q, r and s, you should start by finding all factor pairs of the leading coefficiant and constant term. </span>
the numbers inside the () are first number is x second number is y so it is (x,y)
so for y<x the second number needs to be smaller than the first number
so right away (3,3) is not the answer
and then y also has to be greater then 2x-3
so replace x with the fist numbers and see what you get.
(2x2)-3 = 4-3 = -1 since -1 is not greater than -1 the second answer (2,-1) is not right
3rd one (-1,-2) 2(-1)-3 = -2-3=-5
so -2 is greater than -5 and -2 is less than -1 so the 3rd one is the answer
