HELPP!! Select the correct answer. What is the value of arcsin ?
2 answers:
Answer:
Choice B
Step-by-step explanation:
An option is to find the the square root of 2 in decimals is
Now we can use the arc sine, which is the inverse of a sin.
To do this we must use a scientific calculator. By pressing the arc sin button and entering in 0.7071067812, we can find the arc sin, which is 45°.
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<h3>Answer: Rhombus</h3>
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Reason:
Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.
Segment AB is exactly units long, which is approximately 4.1231 units.
If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.
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Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.
Let's find the slope of AB.
Segment AB has a slope of 1/4.
Do the same for BC
Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.
We don't have perpendicular lines, so the interior angles are not 90 degrees each.
Therefore, this figure is not a rectangle and by extension it's not a square either.
The best description for this figure is a <u>rhombus</u>.
Answer:
18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144
b. The maximum height = 400 feet
c. Attached graph
19) The rocket will reach the maximum height after 4 seconds
20) The rocket hits the ground after 9 seconds
Step-by-step explanation:
* Lets study the rule of motion for an object with constant acceleration
# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time
and a is the acceleration of gravity
# The vertical distances h in x second is h(x) - h(0), where h(0)
is the initial height of the object above the ground
∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial
velocity, a is the acceleration of gravity (32 feet/second²) and x
is the time
18)a.
∵ The value of a = -32 ft/sec² ⇒ negative because the direction
of the motion
is upward
∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16
∴ h(x) = vx - 16x² + h(0)
∴ h(x) = -16x² + vx + h(0) ⇒ proved
* Find the height of the object after x seconds from the ground
∵ h(0) = 144 and v = 128 ft/sec
∴ h(x) = -16x² + 128x + 144
b.
* At the maximum height h'(x) = 0
∵ h'(x) = -32x + 128
∴ -32x + 128 = 0 ⇒ subtract 128 from both sides
∴ -32x = -128 ⇒ ÷ -32
∴ x = 4 seconds
- The time for the maximum height = 4 seconds
- Substitute this value of x in the equation of h(x)
∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet
c. Attached graph
19)
- The object will reach the maximum height after 4 seconds
20)
- When the rocket hits the ground h(x) = 0
∵ h(x) = -16x² + 128x + 144
∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16
∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x
∵ x² - 8x - 9 = 0
∴ (x - 9)( x + 1) = 0
∴ x - 9 = 0 OR x + 1 = 0
∴ x = 9 OR x = -1
- We will rejected -1 because there is no -ve value for the time
* The time for the object to hit the ground is 9 seconds
