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2 years ago

ASAP!!!!!!!!! PLEASE help me with this question! This is really urgent! No nonsense answers please.

Mathematics

1 answer:

Vesna [10]2 years ago

4 0

Answer:

Because <CBD is an inscribed angle and <CAD is a central angle with the same intercepted arc, m<CBD = 55°, or half of the measure of <CAD.

Step-by-step explanation:

The Inscribed Angle Theorem proves that an inscribed angle is half the measure of a central angle, if both the inscribed angle and the central angle intercepts the same arc.

Also, according to the inscribed angle theorem, an inscribed angle is ½ of the measure of the arc it intercepts.

Therefore, m<CBD is half of m<CAD, or half of the measure of the arc CD that they both intercept together.

Thus, m<CBD = 55°, which is ½ of m<arc CD.

m<arc CD = 110° = m<CAD.

m<CBD = ½ of m<CAD = 55°.

The statement that best describes the relationship between <CBD and <CAD is "Because <CBD is an inscribed angle and <CAD is a central angle with the same intercepted arc, m<CBD = 55°, or half of the measure of <CAD."

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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

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

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

Which gives;

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2022} \right)$

$\displaystyle A = 2 \times 1011 \cdot \left(1 - \frac{1}{2022} \right) = \frac{1032231}{511} \approx \mathbf{2020.022}$

A ≈ 2020.022

Learn more about the sum of a series here:

brainly.com/question/190295

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