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sp2606 [1]
3 years ago
10

I beed help nowww hjbiuhkju uhbjknjhiuy

Mathematics
2 answers:
telo118 [61]3 years ago
7 0

Answer:

answer hoga 3/2 ok.....

zalisa [80]3 years ago
5 0

Answer: 3/2

Step-by-step explanation:

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A grocery store sells snacks by weight 6 ounce bag of mixed nuts cost $3.60 predict the cost of two ounce bag
VARVARA [1.3K]

Answer:

1.20

Step-by-step explanation:

you divide the orginal cost by 3

4 0
3 years ago
Read 2 more answers
How do I solve these problems? ln(x) = 5.6 + ln(7.5) and log(x) = 5.6 - log(7.5)
SOVA2 [1]

Use the rules of logarithms and the rules of exponents.

... ln(ab) = ln(a) + ln(b)

... e^ln(a) = a

... (a^b)·(a^c) = a^(b+c)

_____

1) Use the second rule and take the antilog.

... e^ln(x) = x = e^(5.6 + ln(7.5))

... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents

... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms

... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)

2) Similar to the previous problem, except base-10 logs are involved.

... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.

... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5

... x ≈ 53,080.96

8 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
Do 2 or more shapes with the same perimeter have to have the same area? Explain with examples. AND Do 2 or more shapes with the
Murljashka [212]
Yes. Take for example a square and an ellipsis with the same perimeter. The family of ellipses with the same perimeter can have any area between that of a circle to zero (if it is extremely “thin” i.e. if its eccentricity is large). The circle has the maximum area of any other shape with the same perimeter, so the square has the same area of one of the intermediate ellipses.

4 0
3 years ago
Which of the following correctly indicates whether a triangle can have sides with lengths 3.8, 7.4, and 11, and also provides th
dimaraw [331]

Answer:

I think its c?

Step-by-step explanation:

Im sorry if its wrong.

6 0
3 years ago
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