The answer is 2/3 times as heavy as the dog.
Since we are looking for the weight of the cat in relation to the dog's weight, we have to divide the cat's weight by the dog's weight.
5 1/2 = 5.5, 8 1/4 = 8.25.
5.5 / 8.25 = 2/3.
NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by

The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:

Using the quadratic formula, the solution for our problem is

The rocket splashes after 26.845 seconds.
The maximum of this function happens at the root of the derivative. Differentiating our function, we have

The root is

Then, the maximum height is

1029.99 meters above sea level.
Divide I from both sides then subtract r from both sides, too so in the end:
R= (E-r)/I
Given: In the given figure, there are two equilateral triangles having side 50 yards each and two sectors of radius (r) = 50 yards each with the sector angle θ = 120°
To Find: The length of the park's boundary to the nearest yard.
Calculation:
The length of the park's boundary (P) = 2× side of equilateral triangle + 2 × length of the arc
or, (P) = 2× 50 yards + 2× (2πr) ( θ ÷360°)
or, (P) = 2× 50 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)
or, (P) = 100 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)
or, (P) = 100 yards + 209.33 yards
or, (P) = 309.33 yards ≈309 yards
Hence, the option D:309 yards is the correct option.
Answer:
x= -0.01984
Step-by-step explanation:
You divide both sides by 3125 to get x alone.
You get x=-0.01984
Hope this helps!