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1 year ago

7

una palmera de 17 metros de altura se encuentra sujeta por dos cables de 21m y 25m respectivamente. En la figura se pide calcula

r la distancia AB.

1 answer:

aliya0001 [1]1 year ago

8 0

La distancia AB es de 30.66m.

<h3>Cómo calcular la distancia AB?</h3>

Para calcular la distancia AB debemos realizar el siguiente procedimiento matemático utilizando el teorema de Pitágoras como se muestra a continuación.

a² + b² = c²
r² + 17² = 21²
r² = 21² – 17²
r = 12.33m

De acuerdo a lo anterior, podemos establecer que la base del triángulo de la izquierda mide 12.33m. Ahora debemos calcular la distancia de la base del triángulo de la derecha.

a² + b² = c²
s² + 17² = 25²
s² = 25² – 17²
s = 18.33m

De acuerdo a lo anterior, la base del triángulo de la derecha es de 18.33m. Ahora para saber la base total del triángulo que forman ambos cables debemos sumar los valores de la base de los triángulos.

12.33m + 18.33m = 30.66m

Aprenda más sobre sumas en: brainly.com/question/24673650

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Find the area of the rhombus. d1 = 14 m; d2 = 18 m The area of the rhombus is ? m2.

xxTIMURxx [149]

<u><em>Answer:</em></u>

Area of rhombus = 126 m²

<u><em>Explanation:</em></u>

<u>We are given that the two diagonals of the rhombus are:</u>

D₁ = 14 meters and D₂ = 18 meters

<u>The area of the rhombus using its diagonals can be calculated using the following rule:</u>

$Area = \frac{D_1 * D_2}{2}$

<u>Substitute with the given values to get the area as follows:</u>

$Area = \frac{14*18}{2}=126 m^2$

Hope this helps :)

8 0

2 years ago

Read 2 more answers

The claim that 40% of those persons who retired from an industrial job before the age of 60 would return to work if a suitable j

Bad White [126]

Answer:

The p-value of the test is of 0.1922 > 0.02, which means that there is not significant evidence to reject the null hypothesis, that is, there is not significant evidence to conclude that the proportion is of less than 40%.

Step-by-step explanation:

Test if the proportion is less than 40%:

At the null hypothesis, we test if the proportion is of at least 0.4, that is:

$H_0: p \geq 0.4$

At the alternative hypothesis, we test if the proportion is of less than 0.4, that is:

$H_1: p < 0.4$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.4 is tested at the null hypothesis:

This means that $\mu = 0.4, \sigma = \sqrt{0.4*0.6}$

74 out of the 200 workers sampled said they would return to work

This means that $n = 200, X = \frac{74}{200} = 0.37$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.37 - 0.4}{\frac{\sqrt{0.4*0.6}}{\sqrt{200}}}$

$z = -0.87$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.37, which is the p-value of z = -0.87.

Looking at the z-table, z = -0.87 has a p-value of 0.1922.

The p-value of the test is of 0.1922 > 0.02, which means that there is not significant evidence to reject the null hypothesis, that is, there is not significant evidence to conclude that the proportion is of less than 40%.

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2 years ago

What 5/8x + 1/2 (1/4x + 10)?

dedylja [7]

5+3x/4

5x+1/2(1/4+10). Distributive property
5x+x/8+5. Combine like terms
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2 years ago

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Alejandro looked at his first three science tests for the quarter. His scores are represented in different ways. Test 1 number o

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2 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$

$f'(x)=12x^2+26x-12$

$f''(x)=24x+26$

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here: x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

$24(-2.6)+25$

$-37.5$

So we have a maximum at x=-2.6.

$24(.4)+25$

$9.6+25$

$34.6$

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

$y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56$

I'm shoving this into a calculator:

$y=-58.4654411$

So the approximate relative minimum is (0.4,-58.5).

If you graph $y=4x^3+13x^2-12x-56$ you should see the graph taking a dip at this point.

3 0

3 years ago