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VikaD [51]
3 years ago
8

Hey there! I need help me my 6th grade math :-) Last Questions: 8:40 21:17

Mathematics
1 answer:
NeTakaya3 years ago
6 0

3:30 → 1:10 (taking 3 as common)

12:16 → 3:4 (taking 4 as common)

14:28 → 1:2 (taking 14 as common)

15:5 → 3:1 (taking 5 as common)

24:40 → 3:5 (taking 8 as common)

66:11 → 6:1 (taking 11 as common)

2:8 → 1:4 (taking 2 as common)

20:40 → 1:2 (taking 20 as common)

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Plz help I will Give. The Brainliest I don't want my points to be wasted thx.
Genrish500 [490]
C.) because 2/3=.666 then .666/2 = .333 so .333 times 6/5 equals .4 which is the same as 2/5.

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4 years ago
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The value of (-7) 2 is negative. True or false? Please help me!!!! :'(
Katarina [22]

Answer:false

Step-by-step explanation:

(-7)^2

(-7)*(-7)

49

49 is a positive number

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3 years ago
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A company has a profit of$1750 this week. This profit is $900 more than profit p last week. What equals p
den301095 [7]

Answer:

p=$850

Step-by-step explanation:      $900+p=$1750

                              p= $1750-$900=$850

6 0
3 years ago
One of the tallest buildings in a country is topped by a high antenna. The angle of elevation from the position of a surveyor on
irina1246 [14]

Answer:

a. distance of the surveyor to the base of the building = 2051.90 ft

b. height of the building = 1384 ft

c. Angle of elevation from the surveyor to the top of the antenna = 38.31°

d. Height of antenna  =  237.08 ft

Step-by-step explanation:

​The picture above is a illustration of the described event.

a = the height of the flag

b = the height of the building

c = distance of the surveyor from the base of the building

the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°  

distance from her position to the top of the building  = 2475 ft

distance from her position to the top of the flag  = 2615 ft

​(a) How far away from the base of the building is the surveyor​ located?​

using the SOHCAHTOA principle

cos 34° = c/2475

c =  0.8290375726  × 2475

c = 2051.8679921

c = 2051.90 ft

(b) How tall is the​ building

The height of the building = b

sin 34° = opposite /hypotenuse

0.5591929035 = b/2475

b =  0.5591929035  × 2475

b =  1384.0024361

b =  1384.00 ft

​(c) What is the angle of elevation from the surveyor to the top of the​ antenna?

let the angle = ∅

cos ∅ = adjacent/hypotenuse

cos ∅ = 2051.90/2615

cos ∅ =  0.784665392

∅ = cos-1  0.784665392

∅ =   38.310258303

∅ =  38.31°

​(d) How tall is the​ antenna?

height of the antenna = a

sin 38.31° = opposite/hypotenuse

sin 38.31° = (a + b)/2615

sin 38.31° × 2615 = (a + b)

(a + b) =  0.6199159917  × 2615

(a + b) =  1621.0803182

(a + b) = 1621. 08 ft

Height of antenna = 1621. 08 - 1384.00  =  237.08031822 ft

Height of antenna  =  237.08 ft

8 0
3 years ago
SUPER EASY PLS ANSWER!!!!
slamgirl [31]

Answer:3/5

Step-by-step explanation:

7 0
3 years ago
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