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3 years ago

This is another trig question I need help with. Please explain how you do it to me I would appreciate it so much!!

Mathematics

1 answer:

Irina-Kira [14]3 years ago

3 0

Answer:

He stepped back 59.71 feet.

Step-by-step explanation:

Start by making 2 triangles. One has 68 degrees as the acute angle at the base of the triangle and the other has 41 degrees as the acute angle at the base of the triangle. The side opposite of said angle on both triangles will be 80, the height of the tree. You're solving for the bottom (horizontal) side for both triangles.

For the triangle with an acute base angle of 68, the formula to solve for the horizontal side is tan(68) = 80/x.

Get x by itself to get 80/tan(68) = x

x = 32.32.

For the next triangle, the equation is the same but the angle is changed to 41.

tan(41) = 80/x

80/tan(41) = x

x = 92.03

Finally, subtract the smaller distance from the larger:

92.03 - 32.32 = 59.71 steps.

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(g(1) = 50<br> (g(n) = 8 - g(n-1)<br> g(2) =

Finger [1]

G is the same function

Using the 2nd function

g(2) = 8 - g(2-1)
g(2) = 8 - g(1) (g(1)=50 )
So, g(2)= 8-50
g(2) = -42

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3 years ago

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Give three real world examples of rectangular prisms

Sati [7]

Dresser. shoe box. door

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3 years ago

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The length of a rectangle is 2 cm more than four times the width. If the perimeter of the rectangle is 84 cm, what are its dimen

katovenus [111]

Answer:

Length 34 cm

Width 8 cm

Step-by-step explanation:

Let

x-----> the length of the rectangle

y----> the width of the rectangle

we know that

The perimeter of rectangle is equal to

P=2(x+y)

P=84 cm

84=2(x+y)

42=x+y ----> equation A

x=4y+2 -----> equation B

substitute equation B in equation A and solve for y

42=(4y+2)+y

5y=42-2

y=40/5=8 cm

Find the value of x

x=4y+2 ----> x=4(8)+2=34 cm

The dimensions are

Length 34 cm

Width 8 cm

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3 years ago

Identify each measurement is the diameter radius or circumference the tires on a mining truck and 14 feet tall

sukhopar [10]

Answer:

r = 7

c = 43.98

a = 153.94

Step-by-step explanation:

4 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago