I’m having a hard time anyone knows??

A tunnel must be made through a hill. As a result, a surveyor and an engineer create a sketch of the area. The sketch, displayed

Anastasy [175]

Answer:

498 m

Step-by-step explanation:

The AAA theorem states that triangles are similar if all three corresponding angles are equal.

1. Compare triangles FHS and ILS

(a) Reason for similarity

∠F = ∠I = 90°

∠S is common.

∴ ∠H = ∠L

(b) Calculate SL

$\begin{array}{rcl}\dfrac{SF}{SH} & = & \dfrac{SI}{SL}\\\\\dfrac{225}{380} & = & \dfrac{225 + 475}{SL}\\\\225SL & = & 380 \times 700\\& = & 266000\\SL & = & \textbf{1182 m}\\\end{array}$

2. Compare triangles ILS and GLE

(a) Reason for similarity

∠I = ∠G = 90°

∠L is common.

∴ ∠S = ∠E

(b) Calculate LE

$\begin{array}{rcl}\dfrac{IS}{GE} & = & \dfrac{LS}{LE}\\\\\dfrac{700}{180} & = & \dfrac{1182}{LE}\\\\700LE & = & 180 \times 1182\\& = & 212800\\LE & = & \textbf{304.0 m}\\\end{array}$

3. Calculate EH

LE + EH + HS = LS

304.0 m + EH + 380 m = 1182 m

EH + 684 m = 1182 m

EH = 498 m

The distance from E to H is 498 m.

5 0

3 years ago

There are 800 students in the School of Business Administration. There are four majors in the School: Accounting, Finance, Manag

dezoksy [38]

Answer:

Major $X_i$ %

_________________________________________

Accounting 240 (240/800)*100=30%

Finance 160 (160/800)*100=20%

Management 320 (320/800)*100=40%

Marketing 80 (80/800)*100= 10%

_________________________________________

Total 800 100%

Step-by-step explanation:

For this case we can create the following table calculating the percent frequency as:

$\% = \frac{X_i}{n}$

Where $X_i$ represent the number of students for the i category and n the total sample size.

Major $X_i$ %

_________________________________________

Accounting 240 (240/800)*100=30%

Finance 160 (160/800)*100=20%

Management 320 (320/800)*100=40%

Marketing 80 (80/800)*100= 10%

_________________________________________

Total 800 100%

The pie chart is on the figure attached. We use excel to create it.

5 0

2 years ago

Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$