Answer:
vruh
i cant see any shape or image
Step-by-step explanation:
ask me <em>t</em><em>h</em><em>i</em><em>s</em><em> </em><em>a</em><em>g</em><em>a</em><em>i</em><em>n</em><em> </em><em> </em><em>i</em><em>n</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>c</em><em>o</em><em>m</em><em>m</em><em>e</em><em>n</em><em>t</em><em>s</em><em>!</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em><u>i</u></em><em><u>m</u></em><em><u>a</u></em><em><u>g</u></em><em><u>e</u></em><em><u>e</u></em><em><u>e</u></em>
These questions are all of the same type: division. They ask to find how many of a number can fit in another number.
As for question 5, there are 42 pounds of dog food, and
pound of dog food is used for each feeding. To find the number of times Mr. Lash can feed his animals, divide 42 by
. The answer is 48.
Solving the rest of the questions by this method, here are the answers. If you wish you can do them yourself to verify your understanding.
6) 12
7) 4
8) 8
9) 2
Note: in question 9, the question is in reverse order compared to the other questions.
3x-x=(2x+4)-(x+4)? That is my answer but it might not be right
Answer:
4 liters of 60% solution; 2 liters of 30% solution
Step-by-step explanation:
I like to use a simple, but effective, tool for most mixture problems. It is a kind of "X" diagram as in the attachment.
The ratios of solution concentrations are 3:6:5, so I've used those numbers in the diagram. The constituent solutions are on the left; the desired mixture is in the middle, and the numbers on the other legs of the X are the differences along the diagonals: 6 - 5 = 1; 5 - 3 = 2. This tells you the ratio of 60% solution to 30% solution is 2 : 1.
These ratio units (2, 1) add to 3. We want 6 liters of mixture, so we need to multiply these ratio units by 2 liters to get the amounts of constituents needed. The result is 4 liters of 60% solution and 2 liters of 30% solution.
_____
If you're writing equations, it often works well to let the variable represent the quantity of the greatest contributor—the 60% solution. Let the volume of that (in liters) be represented by v. Then the total volume of iodine in the mixture is ...
... 0.60·v + 0.30·(6 -v) = 0.50·6
... 0.30v = 0.20·6 . . . . subtract 0.30·6, collect terms
... v = 6·(0.20/0.30) = 4 . . . . divide by the coefficient of v
4 liters of 60% solution are needed. The other 2 liters are 30% solution.