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3 years ago

5.Simplify each of the following: (i) 7sqrt(2)+5sqrt(8) (ii) 15sqrt(6)-sqrt(216) (iii) 8sqrt(45)-8sqrt(20)+sqrt(245)-3sqrt(125)

Mathematics

1 answer:

emmainna [20.7K]3 years ago

8 0

Answer:

(i) 17 $\sqrt{2}$

(ii) 9 $\sqrt{6}$

(iii) 0

Step-by-step explanation:

(i) 7 $\sqrt{2}$ + 5 $\sqrt{2}$ $\sqrt{4}$ = 7 $\sqrt{2}$ + 10 $\sqrt{2}$ = 17 $\sqrt{2}$

(ii) 15 $\sqrt{6}$ - $\sqrt{216}$ = 15 $\sqrt{6}$ - $\sqrt{36}$ · $\sqrt{6}$ = 9 $\sqrt{6}$

(iii) 8 $\sqrt{45}$ - 8 $\sqrt{20}$ + $\sqrt{245}$ - 3 $\sqrt{125}$ = 8 $\sqrt{9}$ · $\sqrt{5}$ - 8 $\sqrt{4}$ · $\sqrt{5}$ + $\sqrt{49}$ · $\sqrt{5}$ - 3 $\sqrt{25}$ · $\sqrt{5}$ =

24 $\sqrt{5}$ - 16 $\sqrt{5}$ + 7 $\sqrt{5}$ - 15 $\sqrt{5}$ = 8 $\sqrt{5}$ - 8 $\sqrt{5}$ = 0

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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

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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:

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