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3 years ago

108 +621 is the same as a 100 + ? please give answer

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

7 0

Answer:

isn't it 729 , so it is 100+

Step-by-step explanation:

......................

UNO [17]3 years ago

5 0

Answer:

100+629

Step-by-step explanation:

108 + 621

-8 +8

100 + 629

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Answer:

$2^{17}$

Step-by-step explanation:

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3 years ago

In a circle, a chord of 50 cm bisects a chord of 40 cm. Find the length of the longer segment of the 50 cm chord.

Andreyy89

Answer:

<u>The length of the longer segment of the 50 cm chord = 40</u>

Step-by-step explanation:

let the length of the longer segment of the 50 cm chord = x

so, the other segment = 50 - x

A chord of 50 cm bisects a chord of 40 cm

using the Intersecting Chords Theorem

so, x(50-x) = 20 * 20

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x² - 50x + 400 = 0

(x - 40)(x - 10) = 0

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3 years ago

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3 years ago

In order to evaluate 7 sec(θ) dθ, multiply the integrand by sec(θ) + tan(θ) sec(θ) + tan(θ) . 7 sec(θ) dθ = 7 sec(θ) sec(θ) + ta

Maurinko [17]

Answer:

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$

Step-by-step explanation:

The question is not properly formatted. However, the integral of $\int {7 \sec(\theta) } \, d\theta$ is as follows:

$\int {7 \sec(\theta) } \, d\theta$

Remove constant 7 out of the integrand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta$

Multiply by 1

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta$

Express 1 as: $\frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Expand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Let

$u = \sec(\theta) + \tan(\theta)$

Differentiate

$\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)$

Make $d\theta$ the subject

$d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

So, we have:

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

Cancel out $\sec(\theta)\tan(\theta) + sec^2(\theta)$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}$

Integrate

$\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c$

Recall that: $u = \sec(\theta) + \tan(\theta)$

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$

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4 years ago