Answer:
Step-by-step explanation:
A possible word problem for 3x+10 ≤75 could be
Mat has $75 dollars saved in his wallet and wants to use it for go carts. Knowing that the entrance to the park is $10, and each time around the go cart course is $3 what is the maximum times he can go around?
3x+10 ≤ 75, subtract 10 from both sides
3x ≤ 75-10, divide both sides by 3
x ≤ 65/3
or x ≤ 21.(6)
So it can go around 21 times the most.
AA = $1
AAA= $0.75
AA + AAA = 42
$1AA + $0.75AAA= $37
AA + AAA = 42
AA + AAA-AAA= 42- AAA
AA = 42- AAA
$1(42- AAA) + $0.75AAA= $37
$42 - AAA +0.75AAA = $37
$42 -0.25AAA= $37
$42-$42 -0.25AAA= $37 -$42
-0.25AAA= -5
-0.25AAA/-0.25 = -5/-0.25
AAA= 20
AA + AAA= 42
AA + 20 = 42
AA +20 -20 = 42-20
AA= 22
Check
$1AA + $0.75AAA= $37
$1(22)+ $0.75(20)= $37
$22 + $15 =$37
$37 = $37
I am assuming 33 is B^(2/3) times B^(1/2)
See the following image
Answer:
Step-by-step explanation:
a|c means that c=a*k k is some positive integer. We know that b|c so b| ak and (a,b)=1, so it must be b|k, i.e k=b*r, r is some positive integer number. Now we have that c=abr, so ab| c.
B) if x and x’ are both solution then we have that
mi | x-x’ for every i.
By a) we have that m1m2...mk| x-x’, so x and x’ are equal by mod od m1m2...mk.