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3 years ago

Type the correct answer in each box. Consider the expressions shown below.

Mathematics

2 answers:

Korvikt [17]3 years ago

8 0

Answer: 1. A 2. C 3. B

Step-by-step explanation:

pav-90 [236]3 years ago

7 0

Answer:

1 - A, 2 - C, 3 - B

Step-by-step explanation:

The dude above me is right

thanks pal

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Answer8x-2

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4 years ago

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Find the largest possible revenue from the demand equation "Q= -2p + 1000

Mariana [72]

Revenue=quantity x price= (-2p+1000) (p)= -2p^2+1000p
The maximum revenue will occur when the first derivative is zero so when 2(-2p)+1000=0;p=250
Which generates 125,000 in revenue
Try prices of 245 and 255 and you will see they both are less than 250 thereby proving the max revenue is 250

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3 years ago

Create two scenarios that reflect either a markup or markdown in price. For each scenario, write two equations to solve for the

gogolik [260]

Answer:

lucy is buying shoes they are on sale and have been marked down to 8 dollars instead of 20 20-8=12 the shoes were marked down 12 dollars

Brainly is going bankrupt and there making their membership prices higher instead of ten dollars its 15 dollars 15-10=5 they were marked up 5 dollars

Step-by-step explanation:

hope this helps

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3 years ago

Did I do my line to reflect x=1 correctly?

SashulF [63]

Answer:

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Step-by-step explanation:

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

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

8 0

3 years ago