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A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

5 0

3 years ago

A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ

andreyandreev [35.5K]

The correct answer is C) t₁ = 375, $t_n=5t_{n-1}$ .

From the general form,
$t_n=75(5)^n$ , we must work backward to find t₁.

The general form is derived from the explicit form, which is
$t_n=t_1(r)^{n-1}$ . We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

$t_n=t_1(5)^{n-1}$

Using the products of exponents, we can "split up" the exponent:
$t_n=t_1(5)^n(5)^{-1}$

We know that 5⁻¹ = 1/5, so this gives us
$t_n=t_1(\frac{1}{5})(5)^n
\\
\\=\frac{t_1}{5}(5)^n$

Comparing this to our general form, we see that
$\frac{t_1}{5}=75$

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
$t_n=t_{n-1}(r)$ , while we must state what t₁ is; this gives us

$t_1=375; t_n=t_{n-1}(5)$

3 0

3 years ago

Clarissa has a budget of $1,200 a month to spend for rent and food. She has already spent $928 this month. Which inequality repr

shtirl [24]

Answer:

1200-928 = 272 for the rest of the month Clarissa has to spend.

Step-by-step explanation:

So we don't want her to go over her budget so we wouldn't choose a greater than option for her so eliminate the first option and the last option. Now we just have the middle two. We want to keep it under 1200 but theres nothing wrong with spending that exact amount each month so I would pick x less than or equal to 272, since we can still equal the total of less than the total and not go over. Option 2 or B.

6 0

2 years ago

Read 2 more answers

HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP

olga2289 [7]

Answer:can’t read it

Step-by-step explanation:

Can you take a pic closer to the screen?

7 0

3 years ago

Ann Zoe and Jim bought a present for ben. Ann cooperated with $12 less than Zoe and $16 less than Jim. If the cost of the presen

sergeinik [125]

Let A represent the amount Ann contributed, Z represent Zoe's contribution, and J represent Jim's contribution toward the present.

Ann (A): A

Zoe (Z): A = Z - 12 → Z = 12 + A

Jim (J): A = J - 16 → J = 16 + A

A + Z + J = 154

(A) + (12 + A) + (16 + A) = 154

3A + 28 = 154

3A = 126

A = 42

Ann (A): A = 42

Zoe (Z): Z = 12 + A = 12 + 42 = 54

Jim (J): J = 16 + A = 16 + 42 = 58

Answer: Ann=42, Zoe=54, Jim=58

6 0

3 years ago