The area of the square is 81 sq in, and the area of the circule is (3.14)(3 in)^2, or approx 3.14(9) sq in, or approx 28.27 sq in.

Subtract: 81 sq in - 28.27 sq in = approx. 52.73 sq in.

The difference is 52.73 sq in; the square is larger, the circle smaller.

7 0

3 years ago

Costs of Attendance

erastovalidia [21]

Answer:

y = 3,224x + 750

Step-by-step explanation:

Assuming the scholarship and grants are applied using the slope intercept form y=mx+b where m is the slope in this case the cost per year attending x the number of years attended and b the y intercept which in this case is the one time fee subtracting 13,774 by 10,550 gives us 3,224.

Since the one time fee is only in place once that would make it the y intercept therefore y=3,224x+750 is the correct answer

8 0

2 years ago

A car rents for $35 per day plus 18cents¢ per mile. you are on a daily budget of $71. what mileage will allow you to stay with

Butoxors [25]

200 miles
$71-$35=$36 left in budget
$36/.18 per mile= 200 miles

6 0

3 years ago

A bicycle rider increases his speed from 5 m/s to 15 m/s in 10 seconds

valentina_108 [34]

To find the acceleration of the bicycle rider, we are going to use the acceleration formula: $a= \frac{v_{f}-v_{i}}{t}$
where
$a$ is the acceleration
$v_{i}$ is the initial speed
$v_{f}$ is the final speed
$t$ is the time

We know from our problem that increases his speed from 5 m/s to 15 m/s in 10 seconds, so his initial speed is 5 m/s and his final speed is 15 m/s; therefore, $v_{i}=5$ , $v_{f}=15$ , and $t=10$ . Lets replace those values in our formula:
$a= \frac{v_{f}-v_{i}}{t}$
$a= \frac{15-5}{10}$
$a= \frac{10}{10}$
$a=1$

We can conclude that the acceleration of the bicycle rider 1 m/s^2

4 0

3 years ago