All categories

2 years ago

Find the value of x in the given right triangle.

Mathematics

1 answer:

8_murik_8 [283]2 years ago

7 0

Answer:

cosine = adjacent / hypotenuse

cosine (62) = x / 12

x = 12 * 0.46947

x = 5.6

Step-by-step explanation:

You might be interested in

In parts (a) through (e) below, mark the given statement as True or False. Justify each answer. All vectors are in set of rea

Oxana [17]

Answer:

(c)

"The given statement is true, by definition of length of a vector $v$ , $||v|| = \sqrt{v\bullet v}$ "

Step-by-step explanation:

(a) $v \bullet v = || v ||^2$

That is completely correct Remember that if $v = (x_1,x_2,x_3)\\$ then

$v \bullet v = x_1*x_1+x_2*x_2+x_3*x_3 = x_1^2+x_2^2+x_3^2 = ||v||^2$

Therefore the correct answer would be (c).

"The given statement is true, by definition of length of a vector $v$ , $||v|| = \sqrt{v\bullet v}$ "

5 0

3 years ago

Enunciado:Si una empresa determina que sus costos obedecen a un modelo lineal y además sabe que si produce $10$ unidades sus cos

const2013 [10]

Responder:

y = 200x + 2000

Explicación paso a paso:

Deje que la ecuación lineal se exprese como y = mx + c

m es la pendiente

c es la intersección

Si la empresa produce $ 10 unidades, sus costos totales son $ 4000 dólares y si produce $ 15 unidades sus costos ascienden a $ 5000 dólares, entonces podemos representar esto como puntos de coordenadas;

(10,4000) y (15,5000)

Conseguir la pendiente

m = 5000-4000 / 15-10

m = 1000/5

m = 200

Obtener la intersección c

Sustituya m = 2000 y el punto (10, 4000) en la fórmula y = MX + c

4000 = 200 (10) + c

4000-2000 = c

c = 2000

Sustituya m = 200 y c = 2000 en la ecuación y = mx + c

y = 200x + 2000

Por tanto, la expresión lineal requerida es y = 200x + 2000

3 0

3 years ago

√√√√√√√√√√√√√√√√√√√√√√

iogann1982 [59]

Answer:

Step-by-step explanation:

This question is impossible to answer

3 0

2 years ago

Read 2 more answers

What is the answer for 2x-5+6

julsineya [31]

2x - 11 is the answer because you don’t know what x is equal to

5 0

2 years ago

Read 2 more answers

100 POINTS TO RIGHT AWNSER!!!!

makvit [3.9K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586

——————————

The answer is option D) r < 5 or r > – 1.

I'm going to graph each inequality below on a number line.

A) r > 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B) r > 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r$

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C) r < 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r$

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D) r < 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

Now all real numbers are included in the union. So this is the right choice.

Answer: option D) r < 5 or r > – 1.

I hope this helps. =)

6 0

2 years ago