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Rom4ik [11]
2 years ago
7

(trigonometry) solve for x

Mathematics
2 answers:
melamori03 [73]2 years ago
4 0

Answer:

x = 21 \sin(38 \degree)  \\ x = 12.93

Gala2k [10]2 years ago
3 0
<h3>Answer:  17</h3>

The more accurate value is roughly x = 16.5482258257411

=====================================================

Work Shown:

cos(angle) = adjacent/hypotenuse

cos(38) = x/21

21*cos(38) = x

x = 21*cos(38)

x = 16.5482258257411 which is approximate

x = 17

I'm rounding to the nearest whole number since the other side (21) is a whole number. If your teacher instructs you to round otherwise, then follow those instructions of course. Make sure your calculator is in degree mode.

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60+60×0+1 can you solve this for me please
Likurg_2 [28]
60+60×0+1= 61

So your answer is 61.

Hope I Helped!!!
7 0
3 years ago
Use the net to find the lateral area of the prism.
salantis [7]
The lateral area is represented in the diagram by a rectangle 11 cm wide and
(8 + 15 + 17) cm = 40 cm long.

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6 0
2 years ago
a class with 30 students had an average score of 80 on a test. A class with 20 students had an average score of 90 on the same t
snow_tiger [21]

The average score of all the students in both classes is 84.

<h3>How to calculate the average?</h3>

The class with 30 students had an average score of 80 on a test. The total score will be:

= 30 × 80

= 2400

A class with 20 students had an average score of 90 on the same test. The total score will be:

= 20 × 90

= 1800

Total scores = 2400 + 1800 = 4200

Number of students = 20 + 30 = 50

The average score will be:

= Total score / Total students

= 4200 / 50

= 84

Learn more about average on:

brainly.com/question/24313700

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8 0
1 year ago
What is the volume of a cylinder with a radius of 4cm and a height of 8cm
MissTica

Answer: V=πr^2h=π·4^2·8≈402.12386cm^2

6 0
3 years ago
Read 2 more answers
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
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