A pistol is accidentally discharged vertically upward at a height of 3 feet
above the ground. If the bullet has an initial muzzle velocity of 200 feet
per second the path of the bullet is modeled by
h (t) = -16t² + 200t + 3.
A. How long does it take for the bullet to reach its maximum height.
Round to nearest hundredth.
B. What is the maximum height?
C. How long will it take for it to hit the ground?
D. When will the bullet be 403 feet above the ground?
Answers
The maximum height the bullet will reach is 627 ft. The time(from the point when bullet was fired) it will take the bullet to hit the ground is 12.51 sec approx. The time when the bullet will be 403 ft above the ground are 2.475 sec and 10.02 sec approx.
How to obtain the maximum value of a function?To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.
For this case, the height of the bullet from the ground is modeled by the equation:
[tex]h(t) = -16t^2 + 200t + 3[/tex]
where height is in meters and time is in seconds. And t is the time in seconds passed from the point of time when the bullet was fired.
A) Finding maximum height reached by the bullet:Finding its first and second rate with respect to the variable 't', we get:
[tex]\dfrac{d(h(t))}{dt} = h'(t) = -32t + 200\\\\\dfrac{d^2(h(t))}{dt^2} = h''(t) = -32 < 0[/tex]
Thus, the second rate is negative no matter what the value of 't' is. So all the critical points would be corresponding to maxima.
Now, finding the critical points by equating the first rate = 0, we get:
[tex]h'(t) = 0\\\\-32t + 200 = 0\\\\t = \dfrac{200}{32} = 6.25 \: \rm sec.[/tex]
Due to only one critical point, and that each critical point is maxima, we have the value of h(t) globally maximum when t = 6.25
Putting t = 6.25 in the function h(t), we get the maximum height achieved by the bullet as:
[tex]h(t) = -16t^2 + 200t + 3\\h(6.5) = -16(6.5)^2 + 200(6.5) + 3 = 627 \: \rm ft[/tex]
B) Finding the time it will take the bullet to fall on the groundWhen the bullet will fall on the ground, the value of h(t) would be 0.
Putting h(t) = 0, and finding the values of 't' for which this is true, we get:
[tex]h(t) = -16t^2 + 200t + 3\\0 = -16t^2 + 200t + 3\\\\16t^2 -200t -3 =0\\\\t = \dfrac{-(-200) \pm \sqrt{ (-200)^2 - 4(16)(-3)}}{2(16)}\\\\t \approx \dfrac{200 \pm 200.479}{32}\\\\t \approx -0.0149 \: \rm sec., t \approx 12.51 \: sec.[/tex]
Bullet will fall only after it is shot, so time taken for the bullet to fall compared to the time when its shot would be greater, so time passed (t here) would be positive.
Thus, at approx t = 12.51 sec, the bullet will fall on the ground.
C) When will the bullet be at 400 ft of height from the ground?At t = 0, bullet was at 3 ft, and therefore, whenever bullet will reach 400 ft, the time t would be > 0.
Putting h(t) = 400, we get:
[tex]h(t) = -16t^2 + 200t + 3\\400 = -16t^2 + 200t + 3\\\\16t^2 -200t +397 =0\\\\t = \dfrac{-(-200) \pm \sqrt{ (-200)^2 - 4(16)(397)}}{2(16)}\\\\t \approx \dfrac{200 \pm 120.797}{32}\\\\t \approx 2.475\: \rm sec., t \approx 10.02\: sec.[/tex]
These both values are true. The time when t = 2.475 approx, the bullet would be going up, passing by 400 ft height, and when t = 10.02 approx, the bullet would be coming down, passing by 400 ft height.
Thus, the maximum height the bullet will reach is 627 ft. The time(from the point when bullet was fired) it will take the bullet to hit the ground is 12.51 sec approx. The time when the bullet will be 403 ft above the ground are 2.475 sec and 10.02 sec approx.
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Related Questions
2
Find the length of the missing side. Choose the correct answer in simplest radical form.
7
Sin
8 in
7mn
Not drawn to scale
А
V17 m
B
113 m
113 m
m
с
D
71 m
Answers
Answer:
C
Step-by-step explanation:
using Pythagoras' identity in the right triangle.
the square on the hypotenuse is equal to the sum of the squares on the other 2 sides.
let x be the hypotenuse , then
x² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )
x = [tex]\sqrt{113}[/tex]
Luca filled four jars with sweet tea. How
much sweet tea did he have in total?
Answers
Answer:
167.283525618[tex]in^{3}[/tex] or 53.248[tex]\pi[/tex][tex]in^{3}[/tex]
Step-by-step explanation:
first divide the diameter by 2 to get the radius and you'll get 1.6 and since the formula for finding the volume of a cylinder is [tex]\pi r^{2} h[/tex] you then square 1.6 and get 2.56 multiply that by the height and get 13.312 then multiply by pie to get 41.8208814046 and then multiply that by the number of jars you have which is four to get the total 167.283525618
or 53.248[tex]\pi[/tex] the problem did not specify how it wanted the answer to be
the answer is cubed because the volume is 3 dimentional
A school car wash charged $5 for a car and $6 for a van. A total of 86 cars and vans were washed on
the weekend and they earned $475. How many vans and cars were washed?
Answers
Answer:
car and van 602
Step-by-step explanation:
brain liest me please need
There are 427 vans and 435 cars were washed.
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
A mathematical equation is a statement with two equal sides and an equal sign in between. An equation is, for instance, 4 + 6 = 10. Both 4 + 6 and 10 can be seen on the left and right sides of the equal sign, respectively.
We are given that school car wash charged $5 for a car and $6 for a van.
The total of 86 cars and vans were washed on the weekend and they earned $475.
The equation are;
x + y = 86
5x + 6y = 475
Now we have;
5x + 6y = 475
5(86- y) + 6y = 475
y = 435
Therefore,
435 + x= 8
x = - 427
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Can someone help me with this please and thank you!
Answers
Answer:
A
Step-by-step explanation:
I had this same question on a quiz the other day
An acute triangle has two sides measuring 8 cm and 10 cm. What is the best representation of the possible range of values for the third side, s?
2 < s < 18
6 < s < 12.8
s < 2 or s > 18
s < 6 or s > 12.8
Answers
Answer:
your second choice
Step-by-step explanation:
Answer:
B. 6 < s < 12.8
Step-by-step explanation:
I just took the test
Find the measures of angles CFE and DEF.
Answers
The measure of the angle ∠DEF is 85 degrees and the measure of the angle ∠CFE is 57 degrees.
What is a rectangle?It is a polygon that has four sides. The sum of the internal angle is 360 degrees.
In a cyclic quadrilateral, the sum of opposite angles is 180°.
∠DCE + ∠DEF = 180°
∠DEF = 180° - ∠DCF
∠DEF = 180° - 95°
∠DEF = 85°
Similarly for the other two angles, we have
∠CDE + ∠CFE = 180°
∠CFE = 180° - ∠CDE
∠CFE = 180° - 123°
∠CFE = 57°
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How do I solve this equation for n? (6x2n)÷8=15
Answers
Answer:
n = 10
Step-by-step explanation:
12n ÷ 8 = 15
12n = 15 × 8
12n = 120
n = 120/ 12
n = 10
(6x2n) ➗8=15
6x2n= 15
8
2n=15x8/6
2n=5x4
2n=20
n=20/2
n=10
Make f the subject of 12k^2 = root (f+6)/2
Answers
Answer:
f=2×12k⁴-6
Step-by-step explanation:
root(f+6/2)=12k²
take the root away f+6/2=12k²
f+6/2=12k⁴
f=2×12k⁴-6
NEED HELP ASAP thank you :).
Answers
Answer:
1. angle addition postulate
2. angle addition postulate
3. given
4. substitution
5. given
6. substitution
7. subtraction property of equality
4. An object is at rest if all forces
acting on the object have a net
force of zero. If an object has a
force of -5.5 Newtons applied to
it, what force needs to be applied
in order for the object to be at
rest?
Answers
Answer:
+5.5
Step-by-step explanation:
Lets say you have an apple and you apply -5.5 newtons of force to it, for it to be balanceds you have to push the opposite, which is +5.5 newtons.
The giant circle challenge is finally here let’s all work together for this one
Answers
The measure of the angles are: ∠1 = 30°, ∠2 = 30°, ∠3 = 21°, ∠4 = 39°, ∠5 = 39°, ∠6 = 24°, ∠7 = 24°, ∠8 = 66°, ∠9 = 66°, ∠10 = 21°, ∠11 = 60°, ∠12 = 120°, ∠13 = 60°, ∠14 = 120°, ∠15 = 48°, ∠16 = 105°, ∠17 = 27°, ∠18 = 129°, ∠19 = 51°, ∠20 = 129° and ∠21 = 51°
How to determine the measure of the angles?The given parameters are:
AB = 78°
FE = 105°
ED = 27°
CD = 42°
Considering the semicircle ABCD, the measure of arc BC is:
BC = 180 - AB - CD
This gives
BC = 180 - 78 - 42
Evaluate
BC = 60°
Considering the semicircle A FED, the measure of arc A F is:
A F = 180 - FE - ED
This gives
A F = 180 - 105 - 27
Evaluate
A F = 48°
The angle 16 subtends arc FE.
So, we have:
∠16 = FE
This gives
∠16 = 105°
Similarly, the angles 15 and 17 subtend arcs A F and ED respectively.
So, we have:
∠15 = 48°
∠17 = 27°
Angles at the circumference are half the angles at the arc.
This means that:
∠1 = Arc BC/2
∠3 = Arc CD/2
∠4 = Arc AB/2
∠5 = Arc AB/2
∠6 = Arc A F/2
∠10 = Arc CD/2
So, we have:
∠1 = 60/2 = 30°
∠3 = 42/2 = 21°
∠4 = 78/2 = 39°
∠5 = 78/2 = 39°
∠6 = 48/2 = 24°
∠10 = 42/2 = 21°
The angle in a semicircle is 90 degrees.
This means that:
∠2 = 90°
The sum of angles in a triangle is 180 degrees.
This means that:
∠11 = 180 - ∠1 - ∠2
∠12 = 180 - ∠3 - ∠4
So, we have:
∠11 = 180 - 30 - 90 = 60°
∠12 = 180 - 21 - 39 = 120°
Vertical angles are equal.
So, we have:
∠13 = ∠11
∠14 = ∠12
This gives
∠13 = 60°
∠14 = 120°
The sum of angles in a triangle is 180 degrees.
This means that:
∠18 = 180 - ∠17 - ∠6
So, we have:
∠18 = 180 - 27 - 24 = 129°
Vertical angles are equal.
So, we have:
∠20 = ∠18
This gives
∠20 = 129°
Angle on a straight line equals 180°.
So, we have:
∠21 = 180 - ∠18
This gives
∠21 = 180 - 129° = 51°
Vertical angles are equal.
So, we have:
∠19 = ∠21
This gives
∠19 = 51°
The sum of angles in a triangle is 180 degrees.
This means that:
∠7 = 180 - ∠21 - ∠16
So, we have:
∠7 = 180 - 51 - 105
∠7 = 24°
The angle in a semicircle is 90 degrees.
This means that:
∠7 + ∠8 = 90°
So, we have:
24 + ∠8 = 90°
Subtract 24 from both sides
∠8 = 66°
The sum of angles in a triangle is 180 degrees.
This means that:
∠8 = 180 - ∠15 - ∠9
So, we have:
∠8 = 180 - 48 - 66
So, we have:
∠8 = 66°
Hence, the measure of the angles are:
∠1 = 30°, ∠2 = 30°, ∠3 = 21°, ∠4 = 39°, ∠5 = 39°, ∠6 = 24°, ∠7 = 24°, ∠8 = 66°, ∠9 = 66°, ∠10 = 21°, ∠11 = 60°, ∠12 = 120°, ∠13 = 60°, ∠14 = 120°, ∠15 = 48°, ∠16 = 105°, ∠17 = 27°, ∠18 = 129°, ∠19 = 51°, ∠20 = 129° and ∠21 = 51°
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What is equivalent to 3 gallons?
f 4 quarts
g 6 quarts
h 3 quarts
j 12 quarts
Answers
Answer:
6g quarts
Step-by-step explanation:
three is a factor and multiple of 6
Vince worked 705 hours in 15 weeks. At what rate did he work in hours per week? A. 48 hours per week B. 49 hours per week C. 46 hours per week D. 47 hours per week
Answers
Answer:
c
Step-by-step explanation:
Answer: 47
Step-by-step explanation:
Rate in hours /week = 705/ 15
Please i need some help on this question
Answers
Answer:
B
Step-by-step explanation:
It made more machines per hour than the other ones
A gift shop sells 140 wind chimes per month at $90 each. The owners estimate that for each $5 increase in price, they will sell 7 fewer wind chimes per month. Find the price per wind chime that will maximize revenue.
Answers
Answer: $95
Step-by-step explanation:
Te new price will be 90 + 5x, if the price increases "x" times
The number of wind chimes sold per month will become 140 - 7x
[tex]\begin{aligned}&\text {Revenue, } \mathrm{R}(\mathrm{x})=(90+5 \mathrm{x})(140-7 \mathrm{x}) \\&R^{\prime}(x)=5(140-7 x)-7(90+5 x) \\&R^{\prime}(x)=700-35 x-630-35 x \\&R^{\prime}(x)=70-70 x=0 \\&70 x=70 \\&x=1\end{aligned}[/tex]
Therefore, if the price becomes (90 + 5(1)) = $95 per wind chime, then the revenue will be maximum
Find the height of the cylinder. V=271.4 in^3
Answers
Answer:
c. 14.4
Step-by-step explanation:
when your divide 271.4^3 by 1000 you get 7.2 and then you multiply that by 2 and get 14.4 as your answer.
A hexagon with an apothem of 14.7 inches is shown. a regular hexagon has an apothem of 14.7 inches and a perimeter of 101.8 inches. what is the area of the hexagon? square inches
Answers
The area of the considered regular hexagon which has got 14.7 inches of apothem and a perimeter of 101.8 inches is 748.2 sq. inches.
What is apothem?Apothem for a regular polygon is a line segment which originates from the center of the regular polygon and touches the mid of one of the sides of the regular polygon. It is perpendicular to the regular polygon's side it touches.
Regular polygons have all side same and that apothem bisects the side in two parts, (provable by symmetry).
Consider the diagram attached below.
The area of the regular hexagon considered = 6 times (area of triangle ABC) (because of symmetry).
Also, we have:
Area of triangle ABC = 2 times (Area of triangle ABD).
Thus, we get:
Area of the considered hexagon = 6×2×(Area of triangle ABD)
Area of the considered hexagon = 12×(Area of triangle ABD)
Perimeter of a closed figure = sum of its sides' lengths.
There are 6 equal sides in a regular hexagon (due to it being regular).
Thus, if each side is of 'a' inch length, then:
Perimeter = 6×a inches
[tex]101.8 = 6a\\\\\text{Dividing both the sides by 6, to get 'a' on one side}\\\\a = \dfrac{101.8}{6} \approx 16.967 \: \rm inches[/tex]
This is bisected by the apothem.
Thus, we get:
Length of the line segment BD = |BD| = a/2 ≈ 8.483 inches
Since it is given that the length of the apothem = |AD| = 14.7 inches, therefore, we get:
[tex]\text{Area of ABD} = \dfrac{1}{2} \times \rm base \times height \approx \dfrac{14.7 \times 8.483}{2} \approx 62.35 \: \rm in^2[/tex]
Thus, we get:
Area of the considered hexagon = 12×(Area of triangle ABD)
Area of the considered hexagon [tex]\approx 12 \times 62.35 = 748.2 \: \rm in^2[/tex]
Thus, the area of the considered regular hexagon which has got 14.7 inches of apothem and a perimeter of 101.8 inches is 748.2 sq. inches.
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Answer:
748.23
Step-by-step explanation:
On edge
PLSSS HELP IF YOU TURLY KNOW THISS
Answers
Answer:
B or 5 and 1/2
Step-by-step explanation:
Convert into improper fractionsThis gives 69/8 and 25/869/8 - 25/8 gives 44/8Convert to a mixed numberThis gives 5 and 4/8Simplify 4/8 to 1/2So the answer is BAnswer:
The answer is B. 5 1/2
Step-by-step explanation:
8 5/8 - 3 1/8 = ?
The first thing to check for, is that the fractions have the same denominator!
They do: 8
8 5/8 - 3 1/8 = 5 4/8 reduce
5 1/2.
5 1/2 is your answer!
Suppose that replacement times for washing machines are normally distributed with a mean of 9.4 years and a standard deviation of 2 years. Find the replacement time that separates the top 18% from the bottom 82%.
Answers
Using the normal distribution, it is found that the replacement time that separates the top 18% from the bottom 82% is of 11.23 years.
Normal Probability DistributionThe z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score measures how many standard deviations the measure is above or below the mean. Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.In this problem, the mean and the standard deviation are, respectively, given by [tex]\mu = 9.4, \sigma = 2[/tex].
The desired value is the 82nd percentile, which is X when Z = 0.915, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.915 = \frac{X - 9.4}{2}[/tex]
X - 9.4 = 0.915(2)
X = 11.23
The replacement time that separates the top 18% from the bottom 82% is of 11.23 years.
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Nina needs to create a pond space that has a volume of 134 cubic feet and a depth of 4 feet. She proposes three possible pool designs:
a prism with a square top
an inverted cone with a circular opening
a hemisphere with a circular opening
Calculate the area of exposed water for each of Nina's proposed pool designs.
The area of exposed water for the prism pool is
square feet.
The area of exposed water for the inverted cone pool is
square feet.
The area of exposed water for the hemisphere pool is approximately
square feet
Answers
a. Area of the exposed water of the prism pool: 33.5 ft²
b. The area of exposed water for the cone pool = 102.1 ft²
c. Hemisphere pool = πr² = π(4²) = 50.3 ft²
What is the Volume of an Hemisphere?Volume = (2/3)πr
What is the Volume of a Cone?Volume = 1/3πr²h
What is the Volume of a Prism?Volume = Base area × height
a. Area of the exposed water = area of the square top = base area of the prism
Find base area using, Volume = Base area × height. Thus:
134 = Base area × 4
Base area = 134/4 = 33.5 ft²
Area of the exposed water for the prism pool = 33.5 ft²
b. Find the radius of the cone using, volume = 1/3πr²h.
134 = 1/3πr²(4)
(3)(134) = πr²(4)
402/4π = r²
32 = r²
r = 5.7 ft
The area of exposed water for the inverted cone pool = πr² = π(5.7)² = 102.1 ft²
c. The radius is the depth of the hemisphere pool
The area of the exposed water for the hemisphere pool = πr² = π(4²) = 50.3 ft²
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somebody help me 7u7
Answers
Answer:
105 square feet
Step-by-step explanation:
First of all, find the area of the rectangle (length x width). 7*12 = 84. Now, find the area of the triangle [(base x height)/2]. 6*7 = 42. 42/2 = 21.
Add both of these areas together: 84 + 21 = 105. The area is 105 square feet.
Answer:
105 square feet
Step-by-step explanation:
There are two shapes in this figure, a rectangle and a triangle. We have to find the area of the two shapes separately.
Rectangle Areal x w is the rectangle area formula.
12 x 7 = 84.
The area of the rectangle is 84 square feet
Triangle Area[tex]\frac{l * w}{2}[/tex] is the triangle area formula
[tex]\frac{7* 6}{2}[/tex] = 21
84 + 21 = 105 square feet.
MORE HELP PLEASE!!!!!!!
Answers
Answer:
c. 11.7
Step-by-step explanation:
Distance
√(7 + 3)² + (-1 - 5)²√100 + 36√136≅ 11.7The height in feet of the curved roof of an aircraft hangar can be modeled by y=-0.02x^2+1.6x, where x is the horizontal distance in feet from one wall at ground level. What is the greatest height of the hangar?
Answers
The maximum height is the highest level of height an object can reach. . The greatest height of the hangar is 32feet
How to calculate the maximum height of a function?The maximum height is the highest level of height an object can reach. Given the height in feet of the curved roof of an aircraft hangar can be modeled by y=-0.02x^2+1.6x
The velocity of the aircraft is zero at the maximum height. Therefore:
dy/dx = -0.04x + 1.6 = 0
Determine the value of x
0.04x = 1.6
x = 1.6/0.04
x = 40
Substitute x = 40 into the function to get the greatest height
y=-0.02x^2+1.6x
y=-0.02(40)^2+1.6(40)
y = -32 + 64
y = 32ft
Hence the greatest height of the hangar is 32feet
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Jamal borrowed $15 600 for 6 years from a bank. The annual simple interest rate for the first 3 years is 1.6%. From then onward, the annual simple interest rate is increased to 2%. How much interest will he owe at the end of 6 years?
Answers
Answer:
the interest owed is 1684.8
Step-by-step explanation:
Formula for simple interest =
I = Prt, where P is amount borrowed, r for interest rate and t for time.
Since the first three years is 1.6% interest we can write:
I = 15600 x 0.016 x 3
= 748.8
Then for the following 3 years the interest rate is 2%:
I = 15600 x 0.02 x 3
= 936
Adding the values gives us 1684.8
Please answer this with the answer please no clifferhangers! Possible brainliest
Answers
Answer:
28 tiles in 85 minutes
1 tile in 3 minutes
Step-by-step explanation:
Figure out how long it takes Meg to lay one tile - 50 / 15 - 3.33 minutes per tile
This suggests that Jessica should take less than 3.33 mins to lay one tile for the statement to be correct.
If she lays 20 tiles in 85 minutes, it will take her 85 / 20 = 4.25 minutes per tile, so it can't be this one.
If she lays 24 tiles in 85 minutes, it will take her 85 / 24 = 3.5416 minutes per tile, so it can't be this one.
If she lays 28 tiles in 85 minutes, it will take her 85 / 28 = 3.0357 minutes per tile, so it can be this one.
If she lays 1 tile in 3 minutes, it will take her 3 / 1 = 3 minutes per tile, so it can be this one.
If she lays 1 tile in 4 minutes, it will take her 4 / 1 = 4 minutes per tile, so it can't be this one.
:)
There are 7 seniors, 5 juniors and 4 sophomores on the pep squad. Ms. Williams needs to choose 12 students out of the group to sell spirit buttons during lunch. How many ways can the 12 students be chosen?
Answers
The question is an illustration of combination, and there are 1820 ways to select the 12 students
How to determine the number of selection?The distribution of the students is given as:
Senior = 7
Junior = 5
Sophomore = 4
The total number of students is:
Total = 7 + 5 + 4
Evaluate
Total = 16
To select 12 students from the 16 students, we make use of the following combination formula
Ways = 16C12
Evaluate the expression using a calculator
Ways = 1820
Hence, there are 1820 ways to select the 12 students
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Geometry !! please help I will Mark Brainlist ( image attached)
Answers
Answer:
x=14
Step-by-step explanation:
3x+1 and 43 are both vertical angles so they are equivant.
3x+1=43
3x=42
x=14
The angles pictured are vertical angles so we set them equal to each other and solve for x.
3x + 1 = 43
3x = 42
x = 14
the answer is option #3… 14
I hope this helps and is the BRAINLIEST!!
Good luck with your studies :)
If [tex]x = \sqrt{a^{sin^{-1}t}}[/tex],[tex]y =\sqrt{a^{cos^{-1}t}}[/tex], show that [tex]\frac{dy}{dx}= -\frac{y}{x}[/tex].
Please help & don't spam!
Answers
Step-by-step explanation:
[tex]\sf x = \sqrt{a^{sin^{-1} \ t}}\\\\\\Derivative \ rule:\boxed{\dfrac{d(\sqrt{x})}{dx}=\dfrac{1}{2}*x^{\frac{-1}{2}}=\dfrac{1}{2\sqrt{x}}}[/tex]
[tex]\sf \dfrac{d(\sqrt{a^{sin^{-1} \ t}}}{dt}=\dfrac{1}{2\sqrt{a^{sin^{-1} \ t}}}*\dfrac{d(a^{sin^{-1} \ t})}{dt}\\\\\\Derivative \ rule: \boxed{\dfrac{d(a^{x})}{dx}=log \ a *a^{x}}[/tex]
[tex]\sf = \dfrac{1}{2\sqrt{a^{sin^{-1}} \ t}}*a^{sin^{-1} \ t}* log \ a *\dfrac{d(Sin^{-1} \ t)}{dt}\\\\[/tex]
[tex]Derivative \ rule:\boxed{\dfrac{d(sin^{-1} \ x}{dx}=\dfrac{1}{\sqrt{1-x^2}}}[/tex]
[tex]\sf = \dfrac{1}{2\sqrt{a^{sin^{-1} \ t}}}*a^{Sin^{-1} \ t}*log \ a*\dfrac{1}{\sqrt{1-x^2}}}}}\\\\ = \dfrac{a^{Sin^{-1} \ t}*log \ a}{2\sqrt{a^{sin^{-1} \ t}}*\sqrt{1-x^2}}[/tex]
[tex]\boxed{ \dfrac{a^{sin^{-1} \ t}}{\sqrt{a^{sin^{-1} \ t}}}=\dfrac{\sqrt{a^{sin^{-1} \ t}}*\sqrt{a^{sin^{-1} \ t}}}{\sqrt{a^{sin^{-1} \ t}}} = \sqrt{a^{sin^{-1} \ t}}}[/tex]
[tex]\sf = \dfrac{a^{sin^{-1} \ t}*log \ a}{2\sqrt{1-x^2}}[/tex]
[tex]\sf \dfrac{dy}{dt}=\dfrac{d(a^{cos^{-1} \ t})}{dt}[/tex]
[tex]= \dfrac{1}{2\sqrt{a^{cos^{-1} \ t}}}*a^{cos^{-1} \ t}*log \ a *\dfrac{-1}{\sqrt{1-x^2}}}\\\\\\=\dfrac{(-1)*a^{cos^{-1} \ t}*log \ a}{2*\sqrt{a^{cos^{-1} \ t}}*\sqrt{1-x^2}}[/tex]
[tex]\sf = \dfrac{(-1)*\sqrt{a^{Cos^{-1} \ t}}* log \ a }{2\sqrt{1-x^2}}\\\\[/tex]
[tex]\sf \bf \dfrac{dy}{dx}=\dfrac{dy}{dt} \div \dfrac{dx}{dt}\\[/tex]
[tex]\sf \bf = \dfrac{(-1)*\sqrt{a^{cos^{-1} \ t}}*log \ a}{2*\sqrt{1-x^2}} \ \div \dfrac{\sqrt{a^{sin^{-1} \ t}} *log \ a}{2*\sqrt{1-x^2}}\\\\\\=\dfrac{(-1)*\sqrt{a^{cos^{-1} \ t}}*log \ a}{2*\sqrt{1-x^2}} \ * \dfrac{2*\sqrt{1-x^2}}{\sqrt{a^{sin^{-1} \ t}} *log \ a}\\\\= \dfrac{(-1)* \sqrt{a^{cos^{-1} \ t}} }{\sqrt{a^{sin^{-1} \ t}}}\\\\= \dfrac{-y}{x}[/tex]
[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Let's solve ~
[tex]\qquad \sf \dashrightarrow \: x = \sqrt{ {a}^{sin {}^{ - 1}t } } [/tex]
here, let's differentiate it with respect to t ~
[tex]\sf \dashrightarrow \: \dfrac{dx}{dt} = \dfrac{1}{2 \sqrt{a {}^{sin {}^{ - 1}t } } } \times a {}^{sin {}^{ - 1}t } \sdot ln(a) \times \dfrac{1}{ \sqrt{1 - {x}^{2} } }[/tex]
[tex]\sf \dashrightarrow \: \dfrac{dx}{dt} = \dfrac{ \sqrt{ {a}^{sin {}^{ - 1}t } } \sdot ln(a)}{2 \sqrt{1 - {x}^{2} } } [/tex]
[tex]\sf \dashrightarrow \: \cfrac{dt}{dx} = \dfrac{2 \sqrt{1 - {x}^{2} } }{ \sqrt{a {}^{sin {}^{ - 1} t} \sdot ln(a)} }[/tex]
Smililarly,
[tex]\sf \dashrightarrow \: \dfrac{dy}{dt} = \dfrac{1}{2 \sqrt{a {}^{cos{}^{ - 1}t } } } \times a {}^{cos {}^{ - 1}t } \sdot ln(a) \times \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } }[/tex]
[tex]\sf \dashrightarrow \: \dfrac{dy}{dt} = - \dfrac{ \sqrt{ {a}^{cos {}^{ - 1}t } } \sdot ln(a)}{2 \sqrt{1 - {x}^{2} } }[/tex]
Now : Lets get Required result ~
[tex]\sf \dashrightarrow \: \dfrac{dy}{dx} = \dfrac{dy }{dt} \times \dfrac{dt}{dx} [/tex]
[tex]\sf \dashrightarrow \cfrac{dy}{dx} = - \dfrac{\sqrt{ {a}^{cos {}^{ - 1}t } \sdot \cancel{ ln(a)}}}{ \cancel{2 \sqrt{1 - {x}^{2}}}} \sdot \dfrac{ \cancel{2 \sqrt{1 - {x}^{2}} } }{ \sqrt{a {}^{sin {}^{ - 1} t} }\sdot \cancel{ln(a)}}[/tex]
[tex]\sf \dashrightarrow \cfrac{dy}{dx} = - \dfrac{\sqrt{ {a}^{cos {}^{ - 1}t } }}{ \sqrt{a {}^{sin {}^{ - 1} t} }}[/tex]
[tex]\sf \dashrightarrow \cfrac{dy}{dx} = - \dfrac{y}{x} [/tex]
[ since y = [tex]\sf{\sqrt{a^{cos^{-1}t}} } [/tex] and x = [tex]\sf{\sqrt{a^{sin^{-1}t}} } [/tex] ]
Solve for X, please.
Answers
Answer:
x = - 11
Step-by-step explanation:
[tex]\frac{-x+10}{3}[/tex] = 7 ( multiply both sides by 3 to clear the fraction )
- x + 10 = 21 ( subtract 10 from both sides )
- x = 11 ( multiply both sides by - 1 )
x = - 11
An airplane is flying 4,000 feet above the ground. It is approaching the runway. A person is standing on the runway looking up at the plane. If the angle of elevation is 12 degrees
what is the distance of the plane from the runway, to the nearest tenth of a foot?
Answers
i am from afghanistan and you all gus
The trigonometric relation is solved and distance of the plane from the runway is D = 18,818.52 feet
What are trigonometric relations?Trigonometry is the study of the relationships between the angles and the lengths of the sides of triangles
The six trigonometric functions are sin , cos , tan , cosec , sec and cot
Let the angle be θ , such that
sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent
tan θ = sin θ / cos θ
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
Given data ,
An airplane is flying 4,000 feet above the ground
A person is standing on the runway looking up at the plane. And the angle of elevation is 12°
So , from the trigonometric relation , we get
tan 12° = 4000 / D
On simplifying , we get
D = 4000 / 0.21255656167
D = 18,818.52 feet
Hence , the distance is D = 18,818.52 feet
To learn more about trigonometric relations click :
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