Answer:
x₁ > x₂
Step-by-step explanation:
Both actions imply a parable trajectories, since both are projectile shot cases.
Let´s call x₁ maximum distance in the first case
The maximum height is just in the middle of the curve, therefore x₁ the maximum horizontal distance is equal to 60 feet.
In the second case, the parable curve is modeled by:
y = x₂*( 0.08 - 0.002x₂) or y = 0.08*x₂ - 0.002*x₂²
A second degree equation, solving for x₂ and dismissing the value x₂ = 0
we get:
y = 0 ⇒ x₂*( 0.08 - 0.002x₂) = 0 x₂ = 0
And 0,08 - 0.002*x₂ = 0
- 0.002*x₂ = - 0.08
x₂ = 0.08/0.002
x₂ = 40 f
Then x₁ > x₂
Answer:
7 - 16 = 9 x 4 = 36
Step-by-step explanation: 1 - 16 7 times. 9 + 9 + 9 + 9 = 36
You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.
Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.
Respectively the solution for this second figure is 5 sections as well.
The answer is a because it has a negative slope and to find it i did rise over run it rises one block for every four it goes over
Answer:
The length of the flagpole be 41.2 ft .
Option (A) is correct .
Step-by-step explanation:
As given
The angle looking up at the sun is 70°.
A flagpole casts a shadow of 15 ft .
Now by using the trignometric identity .
BC = 15 ft
tan 70° = 2.75 (Approx)
AB = 2.75 × 15
= 41.2 ft
Therefore the length of the flagpole be 41.2 ft .
Option (A) is correct .
