Step-by-step explanation:
260 square inches
is the right answer
Answer:
Step-by-step explanation:
for c it would be 1\8
for E it would be 3\8
Answer:
12 units
Step-by-step explanation:
x is to 16 as 9 is to x
x/16 = 9/x cross multiply
x^2 = 144
x =12
Answer:
468.4 meters
Step-by-step explanation:
Find the width of the rectangle.
The area of a rectangle is A=length*width. Substitute the values of length and area and solve for width.
A=length*width
8,400=140*width
60= width
Use the width of the rectangle to find the circumference of the semicircles. Since each semicircle is half of a circle, the perimeter of the two semicircles is equal to the circumference of one circle.
The circumference of a circle is equal to pid, where d is the diameter. The diameter of the semicircle is the same as the width of the rectangle.
So, the diameter is 60 meters. Substitute the diameter into the formula for the circumference and simplify using 3.14 for pi.
≈188.4
(2 times 140)+ 188.4
So, the perimeter of the track is 468.4 meters.
Answer:
If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.
Step-by-step explanation:
Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:
the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.
Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:
When we use this to replace the 't" in the displacement formula, we et:
This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)
And we can also answer the question about time, since given the same initial velocity 
, the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)
