1 of 10 is equivalent to 10% because 1/10 is .10 and .10•100% is 10% so the newspapers advertising is higher
Answer:
a = 4
Step-by-step explanation:
Here, we want to write and solve the given equation
a - 2.5 = 1.5
a = 2.5 + 1.5
a = 4
To check, we simply substitute a = 4
That would be;
4-2.5 = 1.5
This is correct
Answer:
x = 1 , y = 2
Step-by-step explanation:
Solve for substitution
Answer:
circle area = 1963.495 ft²
Step-by-step explanation:
<u>Given</u>
A rectangle 550 ft by 350 ft packed with circles on a rectangular grid
<u>Find</u>
the area of the largest circle that can be packed this way
<u>Solution</u>
The diameter of the circle must be a factor of both 350 and 550. We want the diameter to be the largest such factor. We can write the factors of 350 and 550 as ...
- 350 = 50 × 7
- 550 = 50 × 11
The largest factor common to both numbers is 50, so that will be the diameter of the largest circle that will fit on a square grid.
The area of a circle is commonly given by the formula ...
A = πr² . . . . . where r represents the radius of the circle.
The radius is half the diameter, so will be (50 ft)/2 = 25 ft. And the area of the circle is ...
A = π(25 ft)² = 625π ft² ≈ 1963.495 ft²
The area of one circle is about 1963.495 square feet.
_____
The total number of circles is 11×7 = 77. Their total area is 77×1963.495 ft² ≈ 151,189 ft². This is 151,189/192,400 ≈ 78.54% of the total area of the rectangle.
"Part A: What is the y-intercept of the function, and what does this tell you about the horse? (4 points)" The y-intercept is (0,8), which tells us that at the beginning the horse is 8 miles from the barn.
"Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 3 hours, and tell what the average rate represents. (4 points)" The value of the function at x=1 is 58 and that at x=3 is 158. Thus, the change in the horse's distance from the barn is 158-58, or 100 feet. The time period involved here is 2 sec. Thus, the average rate of change of the horse's position with respect to time is
100 feet
average rate of change = ---------------- = 50 ft/sec
2 sec
If the horse were to move steadily at a fixed rate from 58 feet to 158 feet from the barn, its average rate would be 50 ft/sec.
"Part C: What would be the domain of the function if the horse continued to walk at this rate until it traveled 508 feet from the barn? (2 points)"
Here time begins at x=0 and ends at x=4 sec. Thus, the appropriate domain here is [0,4] sec.
