All categories

2 years ago

Encuentra las medidas que faltan

Mathematics

2 answers:

tia_tia [17]2 years ago

8 0

Answer:

x=49

Step-by-step explanation:

los 3 angulos del triangulo deben de sumar 180

76+55=131

180-131=49

76+55+49=180

Marta_Voda [28]2 years ago

3 0

X=49 la verdad no se

You might be interested in

ILL GIVE BRAINILEST: Write an equation for the triangle and solve the equation to find the value of the variable then find the a

riadik2000 [5.3K]

Answer:

n= 69 degrees

Step-by-step explanation:

total inside measure of any triangle is 180 degrees

180= n+n+42

180= 2n+ 42

2n= 180-42

2n= 138

n=69

7 0

3 years ago

Read 2 more answers

Which equation has solution x = -3?

Lubov Fominskaja [6]

Your question: Which equation has solution x = -3?

I'd like you to know: I couldn't find any answer that exactly answered X=-3, although there is an answer similar to. But if it was accidental, this would be the most <em>correct </em>answer.

Answer: A) 2x - 7 = -1

|Step-by-step explanation|

2x - 7 = -1 = x=3
3x + 8 = 1 = x=-7/3
x + 8 = 10 = x=2
11 (2x-6) = -6 = x=30/11

8 0

3 years ago

What is the best approximation for the perimeter of a semicircle with a dimeter of 12cm?

taurus [48]

Answer is choice B: the circumference of a circle is the same as perimeter and is defined by πd (or pi times diameter) We can cut this in half because we are working with a semicircle: so we have 6π which approximates to the value in answer when using 3.14 for pi Woops: forgot to add the diameter of 12 so we would have 30.84

6 0

3 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

——————————

Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

3 years ago

In the following image, AB is parallel to DC, and BC is a transversal intersecting both parallel lines. The measure of angle ABC

AlexFokin [52]

Answer:

n° = 62°

p° = 62°

q° = 118°

v° = 84°

w° = 138°

Step-by-step explanation:

angle ABC is 118°

m° + 118° = 180

m° = 180° - 118°

m° = 62°

n° = m° = 62° (corresponding angles are equal since AB is parallel to DC, and BC)

p° = n° = 62° (vertical angles are equal)

q° + n° = 180° (linear pair angles)

q° + 62° = 180°

q° = 180° - 62°

q° = 118°

v° + 96° = 180° (linear pair angles)

v° = 180° - 96°

v° = 84°

w° + 42° = 180 (linear pair angles)

w° = 180° - 42°

w° = 138°

8 0

4 years ago