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3 years ago

What is the distance between (4,7), (2,2)

Mathematics

1 answer:

galina1969 [7]3 years ago

7 0

Answer:

d = $\sqrt{29}$

Step-by-step explanation:

<u>Distance formula: d = </u> $\sqrt{(x2 - x1)^2 + (y2-y1)^2}$

<u>Step 1: Plug in and find the distance</u>

d = $\sqrt{(2 - 4)^2 + (2 - 7)^2}$

d = $\sqrt{(-2)^2 + (-5)^2}$

d = $\sqrt{4 + 25}$

d = $\sqrt{29}$

Answer: d = $\sqrt{29}$

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(7-b)+(3b+2)+(7-b)+(3b+2)<br><br><br><br> Plz help me out I’m a lil stressed out with school

ankoles [38]

Answer:

4b+18

Step-by-step explanation:

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3 years ago

Read 2 more answers

2. y= 22 + 9 and y-9= 2x both correctly

Vlad1618 [11]

Answer:

y-9=2x

Step-by-step explanation:

The reason it is y-9=2x is because y=22+9 is y=31 and you have 5 to give so 5=31 which is not correct.

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3 years ago

Based on the sample of 500 people, 42% owned cats. Calculate the test statistic. Round to two decimal places.

Tanzania [10]

Answer:

z= 3.63

z for significance level = 0.05 is ± 1.645

Step-by-step explanation:

Here p = 42% = 0.42

n= 500

We formulate our null and alternative hypotheses as

H0: p= 0.42 against Ha : p> 0.42 One tailed test

From this we can find q which is equal to 1-p= 1-0.42 = 0.58

Taking p`= 0.5

Now using the z test

z= p`- p/ √p(1-p)/n

Putting the values

z= 0.5- 0.42/ √0.42*0.58/500

z= 0.5- 0.42/ 0.0220

z= 3.63

For one tailed test the value of z for significance level = 0.05 is ± 1.645

Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.

4 0

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:

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