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miss Akunina [59]
3 years ago
13

Kenneth Drew a scale drawing of an apartment the scale of the drawing was 2 meters:1 meter the living room is 6 m long in real l

ife how long is the living room in drawing
Mathematics
1 answer:
atroni [7]3 years ago
3 0

Answer:

3 meters

Step-by-step explanation:

Given the scale drawing :

2m : 1m

Length of living room in real life = 6 m long

2 meter = 1 meter

Real life = scale drawing

If length = 6 meter long ;

using cross multiplication :

2 meter = 1 meter

6 meter = x

2x = 6

x = 6/2

x = 3 meter

Length of living room in drawing = 3 meters

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Answer:

cfc

Step-by-step explanation:

cfc

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3 years ago
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A surveyor is measuring the distance across a small lake. He has set up his transit on one side of the lake 130 feet from a pili
Dimas [21]

Answer:

x = 225.16 feet

Step-by-step explanation:

Given that,

The angle between the piling and the pier is 60°.

A surveyor has set up his transit on one side of the lake 130 feet from a piling that is directly across from a pier on the other side of the lake.

We need to find the distance between the piling and the pier. Let the distance be d. Using the trigonometry to find it as follows :

\tan(60)=\dfrac{x}{130}\\\\x=\tan(60)\times 130\\\\x=225.16\ feet

So, the required distance is 225.16 feet.

4 0
3 years ago
Four members of a tennis team equally shared 2 granola bars at the end of practice. how much of a granola bar did they get?
Rama09 [41]
4 divided by 2 is 2. So, 1/2 of one bar goes to each player. 
7 0
3 years ago
22 pts PLEASE HELP ASAP JUST TELL ME THE PROPERTY NAMES I SHOULD WRITE IN THE BLANK SPACES
iogann1982 [59]

Answer:

1. Communative Property

2. Associative Property

3. Zero Property

Step-by-step explanation:

4 0
3 years ago
Find the minimum and maximum of f(x, y, z) = y + 4z subject to two constraints, 2x + z = 4 and x2 + y2 = 1. g
yuradex [85]
L(x,y,z,\lambda_1,\lambda_2)=y+4z+\lambda_1(2x+z-4)+\lambda_2(x^2+y^2-1)

L_x=2\lambda_1+2\lambda_2 x=0\implies\lambda_1+\lambda_2x=0
L_y=1+2\lambda_2y=0
L_z=4+\lambda_1=0\implies\lambda_1=-4
L_{\lambda_1}=2x+z-4=0
L_{\lambda_2}=x^2+y^2-1=0

\lambda_1=-4\implies \lambda_2x=4\implies\lambda_2=\dfrac4x
1+2\lambda_2y=0\implies\lambda_2y=-\dfrac12\implies8y=-x

x^2+y^2=1\implies (-8y)^2+y^2=65y^2=1\implies y=\pm\dfrac1{\sqrt{65}}
y=\pm\dfrac1{\sqrt{65}}\implies x=\mp\dfrac8{\sqrt{65}}
2x+z=4\implies z=4\pm\dfrac{16}{\sqrt{65}}

We have two critical points to consider: \left(-\dfrac8{\sqrt{65}},\dfrac1{\sqrt{65}},4+\dfrac{16}{\sqrt{65}}\right) and \left(\dfrac8{\sqrt{65}},-\dfrac1{\sqrt{65}},4-\dfrac{16}{\sqrt{65}}\right).

At these points, we respectively have a maximum of 16+\sqrt{65} and a minimum of 16-\sqrt{65}.
6 0
3 years ago
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