If the baker made 5 pounds of icing and used 4/9 of the icing to make decorations, then he used:
5 * 4/9 =
= 20/9 =
= 2 2/9 [of icing]
To find out how much he has left over, we just need to substract 2 2/9 from 5:
5 - 2 2/9 =
= 3 - 2/9 =
= 2 7/9
Answer: The baker has <u>2 7/9</u> of the icing left.
Answer:
42 inches square
Step-by-step explanation:
x=6
108/12=9
9-3=6
12+12+9+9=42
First draw sloping up (steep because he’s running)to the right until you hit 20 on the y axis. Next draw down and to the right for 5 yards until 15 on y axis. Next is a flat horizontal line for 3 seconds on x axis. Finally a shallow slope line down (because he is walking) and right until you hit the x axis.
I have seen this question before and I think you meant 10 more dimes than nickels.
We can use substitution to answer this question. The value of a nickel is 5 cents, and we can use the variable n to represent the number of nickels. The value of a dime is 10 cents, and we can use the variable d to represent the number of dimes.
First lets figure out the equations.
.10d+.5n=2.80 (the number of nickels (n) multiplied by .5 will tell us their money value. Same thing for the dimes)
d-n=10 (since there are 13 more dimes than nickels, the number of dimes value (d) minus the number of nickels value (n) will give us 10)
Now lets isolate a variable in one of the equations, preferably the second one because it doesn't have any visible coefficients,
d-n=10
-n=10-d (subtracted the d from both sides)
n=-10+d (made the n positive)
Now that we have the value of n, we can plug it into the other equation.
.10d+.05n=2.80
.10d+.05(-10+d)=2.80 (we replaced the n with the value that we previously got)
.10d-.5+.05d=2.80 (did the multiplication)
.15d-.5=2.80 (combined like terms)
.15d=3.30 (added the .5 to both sides)
d=22 (divided both sides by the .15)
Now that we know that there are 22 dimes and we also know that there are 10 less nickels than dimes, so we can subtract 10 from 22 to get the number of nickels. 22-10=12
d=22
n=12
The first option,
both have to be negative, because they were originally both part of the one fraction, which was all negative.
