SIN x degrees = 8/10

1 answer:

7 0

$\sin(x) = \frac{8}{10} \\ \alpha = \sin {}^{ - 1} ( \frac{8}{10} ) \\ = 53.13010235$
a (alpha) refers to the angle

you'll still get the same answer if you do
sin^-1 4/5

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There are 110 students in the 6th grade band and 120 students in the 7th grade band. Of the students, 60% of the 6th grade band

r-ruslan [8.4K]

Answer:

36 more students of grade 7 went on a trip than students of grade 6

Step-by-step explanation:

60 % of 110 students= 60/100*110= 66 students

85 % of 120 students= 85/100 * 120= 102 students

No of students of grade 7th more than 6th grade students= 102-66= 36

6 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$