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3 years ago

11

the scale of a tower to its scale model is 1 meter 5cm . if the tower is 6 meters by 20 meters what are the dimensions of the sc

ale model?

1 answer:

nikklg [1K]3 years ago

8 0

1 m = 5 cm

6 m = 6 * 5 = 30 cm
20 m = 20 * 5 = 100 cm

dimensions are : 30 cm by 100 cm

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A statue is mounted on top of a 21 foot hill. From the base of the hill to where you are standing is 57feet and the statue subte

AleksandrR [38]

Please find the attached diagram for a better understanding of the question.

As we can see from the diagram,

RQ = 21 feet = height of the hill

PQ = 57 feet = Distance between you and the base of the hill

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$\angle SPR$ =Angle subtended by the statue to where you are standing.

$\angle x=\angle RPQ$ = which is unknown.

Let us begin solving now. The first step is to find the angle $\angle x$ which can be found by using the following trigonometric ratio in $\Delta PQR$ :

$tan(x)=\frac{RQ}{PQ} =\frac{21}{57}$

Which gives $\angle x$ to be:

$\angle x=tan^{-1}(\frac{21}{57})\approx20.22^{0}$

Now, we know that $\angle x$ and $\angle SPR$ can be added to give us the complete angle $\angle SPQ$ in the right triangle $\Delta SPQ$ .

We can again use the tan trigonometric ratio in $\Delta SPQ$ to solve for the height of the statue, h.

This can be done as:

$tan(\angle SPQ)=\frac{SQ}{PQ}$

$tan(7.1^0+20.22^0)=\frac{SR+RQ}{PQ}$

$tan(27.32^0)=\frac{h+21}{57}$

$\therefore h+21=57tan(27.32^0)$

$h\approx8.45 ft$

Thus, the height of the statue is approximately, 8.45 feet.

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3 years ago

The Cubs have won

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First you need to figure out the games the Cubs lost. 11-7=4, so they lost 4 games.

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3 years ago

Read 2 more answers

Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

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<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y {}^{0} \times {y}^{3 + 1} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

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3 years ago