Help me please help please please

Mathematics

1 answer:

Kitty [74]3 years ago

3 0

truth thud ddg trdffg has y said the chart and the story of roger and the other ways of a dream was clicked in a diagrammatic piece and was adlof of a book

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What is −20÷4/5 ? −25 −16 −116 −125

Serhud [2]

= -20 / 4/5
= -20 * 5/4
= -100/4
= -25

In short, Your Answer would be Option A

Hope this helps!

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3 years ago

No links GEOMETRY HELP PLEASE!! WILL MARK BRAINLEIST! 10

Zinaida [17]

Given:

EFGH is a square.

To find:

The $m\angle FHG$ .

Solution:

We know that all interior angles of a square are right angles.

$m\angle EHG=90^\circ$ (Right angle)

The diagonals of square are always the angle bisectors.

FH is a diagonal of the square. So, it bisects the angle EHG.

$m\angle FHG=\dfrac{m\angle EHG}{2}$

$m\angle FHG=\dfrac{90^\circ}{2}$

$m\angle FHG=45^\circ$

Therefore, the measure of angle FHG is 45 degrees.

4 0

3 years ago

A member of the local rocketry club launches her latest rocket from a large field. At the moment it's fuels is exhausted, the ro

Mekhanik [1.2K]

Answer:

A. h(t) = -16t² +240t +544 . . . 0 ≤ t ≤ 17
B. 544 ft . . . t=0 is defined as the time the fuel runs out and ballistic motion begins
C. 1344 ft
D. 1444 ft

Step-by-step explanation:

A. We don't have a standard model for powered rocket behavior. For a given thrust, mass decreases as fuel burns, so acceleration increases. These interacting variables depend on the percentage of the mass that is fuel, the rate at which fuel burns, and the thrust obtained from a given mass of fuel. Likely, the density of the atmosphere and the drag associated with that are involved, as well. Even if these factors are modeled in a simple way, the solution to the equation(s) involved rarely shows up in an elementary physics or math book.

So, we're left with modeling the ballistic motion after the fuel runs out. Here, we usually ignore air drag (or lift), pretending the entire behavior matches that of an object in a vacuum and a uniform gravitational field on a stationary, flat Earth. The equation for that motion is ...

... height = -1/2·g·t² +v₀·t +h₀ . . . . where v₀ and h₀ are the velocity and height at t=0, the time the fuel runs out.

g is the gravitational constant, usually taken to be 32 ft/s² or 9.8 m/s², depending on the units of the other variables.

We are given v₀ = 240 ft/s and h₀ = 544 ft, so the equation of motion is ...

... h(t) = -16t² +240t +544 . . . . ft, for t in seconds.

B. The problem statement tells us the rocket is 544 ft high when the fuel runs out (at t=0). Our model of rocket behavior begins at t=0, which we define as the time the fuel runs out.

C. For t=5, h(5) = -16·5² +240·5 +544 = 1344 . . . ft

D. The graph shows the maximum height to be 1444 ft at t=7.5 seconds.

Analytically, the vertex is found at t=-b/(2a) = -240/(2·(-16)) = 240/32 = 7.5 . . . seconds. h(7.5) = (-16·7.5 +240)·7.5 +544 = 1444 . . . ft.