Hello,
2 cases:
if 1/3*q-5>0 then
|1/3*q-5|=1/3*q-5
1-|1/3*q-5|=-6
1-(1/3*q-5)=-6
1-1/3*q+5=-6
6-1/3*q=-6
-1/3*q=-12
q=36
else
|1/3*q-5|=-(1/3*q-5)
1-|1/3*q-5|=-6
1+(1/3*q-5)=-6
1+1/3*q-5=-6
-4+1/3*q=-6
1/3*q=-2
q=-6
9514 1404 393
Answer:
1000
Step-by-step explanation:
If the number of protesters per minute remains a constant, then you could write the proportion ...
p/12 = 177/2.1
Multiplying by 12 gives ...
p = 12(177/2.1) ≈ 1011.4
Here, minutes are given to 2 significant figures, and the initial count is given to 3 significant figures. The best you can hope for is that your estimate is good to 3 significant figures:
1010 protesters
It is probably sufficient to report the number to 2 significant figures*:
1000 protesters
_____
* Unfortunately, with a number like 1000, the only way you can tell it has 2 significant figures is to report it as 1.0×10³ or 10. hundreds. The trailing zeros are usually not considered significant.
A. 7/11= 7 ÷ 11
B. 5/2= 5 ÷ 2
C. 9/10= 9 ÷ 10
D. 7/15= 7 ÷ 15
A fraction is just a division problem. 7/11 would be considered 7 ÷ 11 (or just 7/11) just do the same for the rest of the fractions.
U can use elimination and multiply all the terms in the top expression to get 3x-3y = -4, or 4x = -4, so x = -1. plug it in to find y if needed
