The <em>correct answers</em> are:
5x²+70x+245 ≥ 1050; and
Yes.
Explanation:
Let x be the width of the tablet. Since the width of the TV is 7 inches more than the tablet, the width of the TV would be x+7.
The length of the TV is 5 times the width; this makes the length 5(x+7) = 5x+35.
The area of the TV would be given by
(x+7)(5x+35).
Since Andrew wants the area to be at least 1050, we set the expression greater than or equal to 1050:
(x+7)(5x+35) ≥ 1050
Multiplying this, we have:
x*5x+x*35+7*5x+7*35 ≥ 1050
5x²+35x+35x+245 ≥ 1050
Combining like terms,
5x²+70x+245 ≥ 1050
To see if 8 is a reasonable width for the tablet, we substitute 8 for x:
5(8²)+70(8)+245 ≥ 1050
5(64)+560+245 ≥ 1050
320+560+245 ≥ 1050
1125 ≥ 1050
Since this inequality is true, 8 is a reasonable width.
Answer:
Step-by-step explanation:
I think option 4 is the correct answer
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D
The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3
Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.
Answer:
A
I Had that question before
