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3 years ago

Hahahahahahhhah no links

Mathematics

2 answers:

blondinia [14]3 years ago

7 0

HEREEEEEE USWKWVQOWYWUWG W

ohaa [14]3 years ago

6 0

Answer:

Step-by-step explanation:

9-3z+4+6z-2

13-2+3z

11+3z

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Which expression is equivalent to 25a + 5b - 13

Ainat [17]

5(5a+b)-13

You would have to distribute

5*5a=25a

5*b=5b

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4 years ago

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Travka [436]

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3 years ago

Find all unique zeros of the function fix) 8x- 64x + 160x- 128-

LenKa [72]

Answer: Lets divide by 8. My god, those coefficients are large.

x^3 - 8x^2 + 20x - 16 = 0

Possible integral roots are: +1, -1, +2, -2, +4, -4, +8, -8

If x = 2, f(x) = 8 - 32 + 40 - 16 = 0

Hence, x-2 is a factor.

Use synthetic division to get: x^2 - 6x + 8 = (x-4)(x-2).

Therefore, there are two integral roots: +2 and +4.

Step-by-step explanation:

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3 years ago

All of the following are defined terms except _____.<br> A.ray<br> B.angle<br> C.plane<br> D.segment

FromTheMoon [43]

A plane is an undefined term. =)

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3 years ago

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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

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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3 years ago