He starts saving $ in January to buy a gift in June. (which is 6 months, or 5 months not including June)
Each month he saves 2/3 of his allowance, which is $14.
[his allowance each month is $21, 2/3 of his allowance is $14]
(you multiply 21 by 2/3 = 42/3 = 14)
The gift he wants to buy is $110
x = the number of months
y = total cost
14x = y
14x = 110 [he saves $14 each month, he wants to have a total of $110 to buy a gift]
Plug in 6 for x in the equation
14(6) = 110
84 = 110 (He is $26 short, so saving either for 5 or 6 months will not get Ian $110)
Ian will not have enough money because he is $26 short for 6 months or $40 short for 5 months. (you can decide whether you want to go by 5 months or 6(including June), sorry for the confusing answer)
Answer:
Explanation:
<u>Description of triangle EFD</u>:
- Angle EFD: right angle (90º)
- x: adjacent leg to angle EDF
- 3: hypotenuse of triangle EFD
<u>Description of triangle EDG</u>:
- Angle EDG: right angle (90º)
- 4 adjacent leg to angle EGD
- 5: hypotenuse of triangle EDG
<u>Similarity</u>:
- Triangle EFD and triangle EGD are similar.
Then, since corresponding parts of similar triangles are proportional:
<u>Clear x</u>:
Answer:
II. One and only one solution
Step-by-step explanation:
Determine all possibilities for the solution set of a system of 2 equations in 2 unknowns. I. No solutions whatsoever. II. One and only one solution. III. Many solutions.
Let assume the equation is given as;
x + 3y = 11 .... 1
x - y = -1 ....2
Using elimination method
Subtract equation 1 from 2
(x-x) + 3y-y = 11-(-1)
0+2y = 11+1
2y = 12
y = 12/2
y = 6
Substitute y = 6 into equation 2:
x-y = -1
x - 6 = -1
x = -1 + 6
x = 5
Hence the solution (x, y) is (5, 6)
<em>Hence we can say the equation has One and only one solution since we have just a value for x and y</em>
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