Let 'w' represent the width of the parking lot, then 'w+6' represents its length
area = width * length
160 = w * (w+6)
solving for 'w' we have a positive value of w=10
10+6=16
the width of the parking lot is 10 yards, the length of the parking lot is 16 yards
Answer:
No
Step-by-step explanation:
11. Correct
11a. a is the number of accessories bought
12. 32 = 3.5a + 25
7 = 3.5a
a = 2
12a. cost of the bear
13. correct
13a. f(-1)=8
2500/10=250
250 represents how much the elephant can eat in one day. Since they want to know how much the elephant can eat in 1000 days you would multiply 250 and 1000 which leaves you with 250,000 pounds of food.
The elephant can eat 250,000 pounds of food in 10 days. :)
Answer:
Step-by-step explanation:
1/4 x 2 = 2/4
2/4 simplified = 1/2
ANSWER

Or

EXPLANATION
The given polynomial is

where a=1,b=k, c=30
Let the zeroes of this polynomial be m and n.
Then the sum of roots is

and the product of roots is

The square difference of the zeroes is given by the expression.

From the question, this difference is 169.
This implies that:





We substitute the values of k into the equation and solve for x.


The zeroes are given by;

