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SCORPION-xisa [38]
3 years ago
11

If a string that is 15.5 inches long is cut into equal pieces that are 1.25 inches long, how many pieces of ribbon can be create

d?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
8 0
Number created  =  15.5 / 1.25  =  12.4
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Solve for x <br><br> Please show your work <br><br> Thank you
Licemer1 [7]
6-4=2
5-X=?
5-2=3
X=3
That is the answer, sorry if i didn't explain it well<span />
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
Write 2 1/16 as a decimal
devlian [24]

Answer:

2.0625

Step-by-step explanation:

2+1/16

= 2.0625

= 206.25%

6 0
3 years ago
Read 2 more answers
200 σ j=1 2j( j 3) describe the steps to evaluate the summation. what is the sum?
ziro4ka [17]

The sum of the equation is  = 5494000.

<h3>What does summation mean in math?</h3>

The outcome of adding numbers or quantities mathematically is a summation, often known as a sum. A summation always has an even number of terms in it. There may be just two terms, or there may be 100, 1000, or even a million. Some summations include an infinite number of terms.

<h3>Briefing:</h3>

Distribute 2j to (j+3).

Rewrite the summation as the sum of two individual summations.

Evaluate each summation using properties or formulas from the lesson.

The lower index is 1, so any properties can be used.

The sum is 5,494,000.

<h3>Calculation according to the statement:</h3>

\sum_{j=1}^{200} 2 j(j+3)

simplifying them we get:

\sum_{j=1}^{200} 2 j^{2}+6 j

Split the summation into smaller summations that fit the summation rules.

\sum_{j=1}^{200} 2 j^{2}+6 j=2 \sum_{j=1}^{200} j^{2}+6 \sum_{j=1}^{200} j

\text { Evaluate } 2 \sum_{j=1}^{200} j^{2}

The formula for the summation of a polynomial with degree 2

is:

\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}

Substitute the values into the formula and make sure to multiply by the front term.

(2)$$\left(\frac{200(200+1)(2 \cdot 200+1)}{6}\right)$$

we get: 5373400

Evaluating same as above : 6 \sum_{j=1}^{200} j

we get: 120600

Add the results of the summations.

5373400 + 120600

= 5494000

The sum of the equation is  = 5494000.

To know more about  summations visit:

brainly.com/question/16679150

#SPJ4

6 0
2 years ago
Can somebody plz help answer these questions (only if u know how to do this) thanks a lot! :)
Mamont248 [21]

Answer:

12. 3:12

Step-by-step explanation:

that is 12

4 0
2 years ago
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