Answer:
<u>Solution for (a)</u>
Earlier Copper pennies are 22.
After 1983 pennies are 28.
Step-by-step explanation:
Pennies minted after 1983 weigh 2.50g
Earlier copper pennies weigh 3.11g
A roll of pennies contains 50 pennies.
A particular roll of pennies weighs 138.42g
Let the number of pennies minted after 1983 be x and;
Earlier copper pennies be 50 - x
2.50x + 3.11(50 - x) = 138.42
Opening the brackets gives;
2.5x + 155.5 - 3.11x = 138.42
Solving like terms together gives;
-0.61x = -17.08
Dividing through gives;
x = 28
So there are 28 after 1983 pennies and;
50 - 28 = 22 earlier pennies.
Answer:
800
Step-by-step explanation:
960 = 2 x 480, so the bacteria will double twice
200 x 2 x 2 = 800
It seems that some the work is already here, but I'd be glad to!! So for #3 which is 9x^2+15x, we can factor out both a 3 and an x (3x) so we know that 3x * 3x =9x^2 and 3x * 5 = 15x so once we take the 3x out of the equation, we are left with 3x(3x+5) and that's as far as you can factor.
For #4, we see that the common factor is 10m because 10m * 2n = 20mn and 10m * 3 = 30m so once we take 10m out of the original, it becomes 10m(2n-3)
For #5, this one the common factor is 4xy because 4xy * 2xy=8x^2y^2 and 4xy*x= 4x^2y and 4xy*3=12xy so once we take the 4xy out of the equation, it becomes 4xy(2xy-x-3)
Hope this helps!
Easy
f(g(1))
evaluate g(1) then plug thatin for x in f(x)
g(1)=(x+2)/3
g(1)=(1+2)/3
g(1)=1
f(g(1))=
f(1)=(1)^2+3(1)+6
f(1)=1+3+6
f(1)=10
f(g(1))=10