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Radda [10]
3 years ago
15

If you place a 23-foot ladder against the top of a building and the bottom of the ladder is 11 feet from the bottom of the build

ing, how tall is the building? Round to the nearest tenth of a foot. PLEASE HURRY

Mathematics
2 answers:
ale4655 [162]3 years ago
5 0

Answer:

12 feet and 2/10

Step-by-step explanation:

zlopas [31]3 years ago
3 0

Answer:20.2

Step-by-step explanation:

a^2+(11)^2 =23^2 a^2 +121=529

a^2=408 a=20.199009…

a=20.2

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Chee is flying a kite, holding her hands a distance of 3.25 feet above the ground and
Mademuasel [1]

The height of the kite above the ground is 58.68 ft

Let x be the height of the kite above Chee's hand.

The height of the kite above Chee's hand, the string and the horizontal distance between Chee and the kite form a right angled triangle with hypotenuse side, the length of the string and opposite side the height of the kite above Chee's hand.

Since we have the angle of elevation from her hand to the kite is 29°, and the length of the string is 100 ft.

From trigonometric ratios, we have

tan29° = x/100

So, x = 100tan29°

x = 100 × 0.5543

x = 55.43 ft.

Since Chee's hand is y = 3.25 ft above the ground, the height of the kite above the ground, L = x + y

= 55.43 ft + 3.25 ft

= 58.68 ft to the nearest hundredth of a foot

So, the height of the kite above the ground is 58.68 ft

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2 years ago
I can't figure out how to do (i + j) x (i x j)for vector calc
Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}

Then, solving the determinant

\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}

In our case,

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Where we used the formula for AxB to calculate ixj.

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\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}

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