Answer:
X = 2
Y = 1
Step-by-step explanation:
x + 4y = 6
3x + 3y = 9
3x + 12y = 18
3x + 3y = 9
9y = 9
x + 4y = 6
y = 1
x + 4 * 1 = 6
y = 1
x + 4 = 6
y = 1
x = 2
so there ia only 1 solution for this equation is (2;1)
Done :))
Answer:
401
Step-by-step explanation:
1. Approach
To solve this problem, one first has to think about the given figure in a certain way. In the figure, one can see that it is a circle attached on either side of a rectangle. To find the perimeter of the figure, one has to find the circumference of the circle and then add two sides of the rectangle to the answer
2. Circumference of the circle
The formula to find the circumference of a circle is;
(pi) or 
(pi)
~ diameter times the value (pi)
Normally to find the circumference of a semicircle, one would have to divide this formula by 2, but since in this case, one has to add two congruent semicircles, so therefore, the effect of dividing the equation by two, only to multiply by two again cancels, and hence, there is no need to divide by 2.
Substitute in the values;
(78)(pi)
~ 245
3. Find the perimeter of the entire object
Now, one has to add the two additional sides of the figure, to the circumferences of the semicircles to get the final answer;
78 + 78 + 245
= 401
Answer: a because it’s using the commutative property of addition meaning it’s the same mathematical statement, it’s just moved around
Step-by-step explanation:
Answer:
x=2
Step-by-step explanation:
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
