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3 years ago

9

A 14 ft ladder is used to scale a 13 ft wall. At what angle of elevation must the laxder be situated in order to reach the top o

f the wall?

2 answers:

Crazy boy [7]3 years ago

7 0

Answer:

The angle of elevation is equal to 68.2 degrees

Step-by-step explanation:

As you can see from the image below we are trying to find the angle on the bottom right.

We know that the hypotenuse is equal to 14 and we know that the opposite is equal to 14. This means that we will be using Sine.

sin x = 13/14

divide:

sin x = 0.9285714286

now we need to use sin^-1

sin^-1 (0.9285714286) = 68.21321071

Round 68.21321071 to the nearest tenth:

68.2

The angle of elevation is 68.2 degrees!

Allisa [31]3 years ago

7 0

Answer:

$\theta \approx 68^\circ$

Step-by-step explanation:

<u>Trigonometry</u>

The ladder, the wall, and the ground form a right triangle, where the ladder is the longest side (hypotenuse).

The trigonometric ratios stand on right triangles and can be used to relate side lengths with angles.

The angle of elevation, formed by the ladder and the ground has the height of the wall as the opposite side. Thus we can use the sine ratio, defined as:

$\displaystyle \sin\theta=\frac{\text{opposite leg}}{\text{hypotenuse}}$

$\displaystyle \sin\theta=\frac{13}{14}$

Solving:

$\displaystyle \theta=\arcsin \frac{13}{14}$

Calculating:

$\mathbf{\theta \approx 68^\circ}$

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Can someone help me with this

lawyer [7]

I believe it is one!

3 0

3 years ago

For each, list three elements and then show it is a vector space.

Butoxors [25]

Answer:

(a) Three polynomials of degree 1 with real coefficients belong to the set $P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}$ , then:

$2+3x \in P_1$

$4.5+\sqrt2 x \in P_1$

$\log5+78x \in P_1$

(b) Three polynomials of degree 1 with real coefficients that hold the relation $a_0 - 2a_1 = 0$ belong to the set $P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}$ . The relation between the coefficients is equivalent to $a_1 = \frac{a_0}{2}$ , then:

$4+2x \in P_2$

$13+6.5x \in P_2$

$10.5+5.25x \in P_2$

Step-by-step explanation:

(a) Three polynomials of degree 1 with real coefficients belong to the set $P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}$ , then:

$2+3x \in P_1$
$4.5+\sqrt2 x \in P_1$
$\log5+78x \in P_1$

A vector space is any set whose elements hold the following axioms for any $\vec{u}, \vec{v}$ and $\vec{w}$ and for any scalar $a$ and $b$ :

$(\vec{u} + \vec{v} )+\vec{w} = \vec{u} +( \vec{v} +\vec{w})$
There is the <em>zero element </em>such that: $\vec{0} + \vec{u} = \vec{u} + \vec{0}$
For all element $\vec{u}$ of the set, there is an element $-\vec{u}$ such that: $-\vec{u} + \vec{u} = \vec{u} + (-\vec{u}) = \vec{0}$
$\vec{u} + \vec{v} = \vec{v} + \vec{u}$
$a(b\vec{v}) = (ab)\vec{v}$
$1\vec{u} = \vec{u}$
$a(\vec{u} + \vec{v} ) = a\vec{u} + a\vec{v}$
$(a+b)\vec{v} = a\vec{v}+b\vec{v}$

Let's proof each of them for the first set. For the proof, I will define the polynomials $a_0+a_1x$ , $b_0+b_1x$ and $c_0+c_1x$ and the scalar $h$ and $g$ .

$(a_0+a_1x + b_0+b_1x)+c_0+c_1x = a_0+a_1x +( b_0+b_1x+c_0+c_1x)\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x$ and defining $a_0+b_0+c_0 = \alpha_0$ and $a_1+b_1+c_1 = \alpha_1$ , we obtain $\boxed{\alpha_0+\alpha_1x= \alpha_0+\alpha_1x}$ which is another polynomial that belongs to $P_1$
A null polynomial is define as the one with all it coefficient being 0, therefore: $\boxed{0 + a_0+a_1x = a_0+a_1x + 0 = a_0+a_1x}$
Defining the inverse element in the addition as $-a_0-a_1x$ , then $-a_0-a_1x + a_0 + a_1x = a_0+a_1x + (-a_0-a_1x)\\\boxed{(-a_0+a_0)+(-a_1+a_1)x = (a_0-a_0)+(a_1-a_1)x = 0}$
$(a_0+a_1x) +( b_0+b_1x) =( b_0+b_1x) +( a_0+a_1x)\\(a_0+b_0)+(a_1+b_1)x = (b_0+a_0)+(b_1+a_1)x\\\boxed{(a_0+b_0)+(a_1+b_1)x = (a_0+b_0)+(a_1+b_1)x}$
$a[b(a_0+a_1x)] = ab (a_0+a_1x)\\a[ba_0+ba_1x] = aba_0+aba_1x\\\boxed{aba_0+aba_1x = aba_0+aba_1x}$
$\boxed{1 \cdot (a_0+a_1x) = a_0+a_1x}$
$\boxed{a[(a_0+a_1x)+(b_0+b_1x)] = a(a_0+a_1x) + a(b_0+b_1x)}$
$(a+b)(a_0+a_1x)=aa_0+aa_1x+ba_0+ab_1x\\\boxed{(a+b)(a_0+a_1x)= a(a_0+a_1x) + b (a_0+a_1x)}$

With this, we proof the set $P_1$ is a vector space with the usual polynomial addition and scalar multiplication operations.

(b) Three polynomials of degree 1 with real coefficients that hold the relation $a_0 - 2a_1 = 0$ belong to the set $P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}$ . The relation between the coefficients is equivalent to $a_1 = \frac{a_0}{2}$ , then:

$4+2x \in P_2$
$13+6.5x \in P_2$
$10.5+5.25x \in P_2$

Let's proof each of axioms for this set. For the proof, I will define again the polynomials $a_0+a_1x$ , $b_0+b_1x$ and $c_0+c_1x$ and the scalar $h$ and $g$ . Again the relation $a_1 = \frac{a_0}{2}$ between the coefficients holds

$[(a_0+a_1x) +( b_0+b_1x)]+(c_0+c_1x) = (a_0+a_1x) +[( b_0+b_1x)+(c_0+c_1x)]\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x$ and considering the coefficient relation and defining $a_0+b_0+c_0 = \alpha_0$ and $a_1+b_1+c_1 = \alpha_1$ , we have $(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x\\(a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x = (a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x\\\boxed{\alpha_0 + \alpha1x = \alpha_0 + \alpha1x}$ which is another element of the set since it is a degree one polynomial whose coefficient follow the given relation.

The proof of the other axioms can be done using the same logic as in (a) and checking that the relation between the coefficients is always the same.

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