All categories

3 years ago

Calculo el area del búmeran tomando en cuenta que su diámetro es 20 cm

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

7 0

Answer:

50π cm²

Step-by-step explanation:

In this case we have that the area of the boomerang has been the area of the largest semicircle minus the area of the smaller semicircles.

We know that the radius is half the diameter:

r = d / 2 = 20/2

r = 10

Now we have to:

Alargest = π · r²

Alargest = π · (10 cm) ²

Alargest = 100π cm²

Asmaller = π · r²

Asmaller = π · (5 cm) ²

Asmaller = 25π cm²

Finally, the boomerang area has been:

Aboomerang = 100π cm² - 2 · (25π cm²)

Aboomerang = 50π cm²

You might be interested in

**ANSWER NEEDED QUICKLY** Here is a number pyramid puzzle.

Schach [20]

Answer:

ready. took me 2 minutes

6 0

2 years ago

There are 57 students in the orchestra and twice that number in the band. There are 18 boys and 28 girls in the choir. If each s

Klio2033 [76]

There are 217 students total

7 0

3 years ago

Read 2 more answers

Factor x2+4x+3.

Alexxx [7]

Answer:

c. (x +3)(x + 1)

Step-by-step explanation:

Factoring an equation is a way of expressing an equation as the product of other linear expressions. The following expression,

$x^2 +4x +3$

Can be represented as the product of the following,

$(x+3)(x+1)$

To check this, one can distribute, one of the expressions over the other, then simplify, the result should be the starting expression,

$(x+3)(x+1)\\$

Distribute,

$(x)(x)+(1)(x)+(3)(x)+(3)(1)$

Simplify,

$x^2 +x +3x + 3\\$

$x^2 +4x +3$

8 0

3 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

3 years ago

I really need help asap please

gizmo_the_mogwai [7]

We can use the substitution method to solve this problem.

The second equation is $\sf y=2x$ , so we can plug in 2x for 'y' in the first equation:

$\sf 5x-2y=3$

$\sf 5x-2(2x)=3$

Multiply:

$\sf 5x-4x=3$

Combine like terms:

$\sf x=3$

This is the x-value of our solution, we can plug this into any of the two equations to find the y-value:

$\sf y=2x$

$\sf y=2(3)$

Multiply:

$\sf y=6$

This is the y-value of our solution. So our entire solution is (3, 6).

5 0

2 years ago