Answer: $ 38.69 ; There are 12 people in the crew
Answer:
Step-by-step explanation:
Let's call hens h and ducks d. The first algebraic equation says that 6 hens (6h) plus (+) 1 duck (1d) cost (=) 40.
The second algebraic equations says that 4 hens (4h) plus (+) 3 ducks (3d) cost (=) 36.
The system is
6h + 1d = 40
4h + 3d = 36
The best way to go about this is to solve it by substitution since we have a 1d in the first equation. We will solve that equation for d since that makes the most sense algebraically. Doing that,
1d = 40 - 6h.
Now that we know what d equals, we can sub it into the second equation where we see a d. In order,
4h + 3d = 36 becomes
4h + 3(40 - 6h) = 36 and then simplify. By substituting into the second equation we eliminated one of the variables. You can only have 1 unknown in a single equation, and now we do!
4h + 120 - 18h = 36 and
-14h = -84 so
h = 6.
That means that each hen costs $6. Since the cost of a duck is found in the bold print equation above, we will sub in a 6 for h to solve for d:
1d = 40 - 6(6) and
d = 40 - 36 so
d = 4.
That means that each duck costs $4.
So,
5y*3 is the open phrase the student uses to model "the sum of 5y and 3".
"The sum of" means addition. The student put 5y*3, while the sum of 5y and 3 is actually 5y + 3.
Answer:
-1ab^2 + 5b + 8
Step-by-step explanation:
3a^2+9ab+5-4a^2-4ab+3
3a^2-4ab^2=-1ab^2
9ab-4ab=5ab
5+3=8
-1ab^2+5ab+8
90°+b+b =180
90°+ 2b=180
2b=180-90
2b= 90
<u>2</u><u>b</u><u>=</u><u>9</u><u>0</u>
<u>2</u> = 2
b=45°
ave tried I don't know if its correct
