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2 years ago

¿se puede hacer un triangulo con las siguientes medidas:2cm ,3cm ,10cm? ayudenme es para mañana

1 answer:

3 0

Não, não podemos fazer um triângulo com os comprimentos dos lados de 2 cm, 3 cm e 10 cm. Isso ocorre porque a soma de 2+3 < 10. (in english: No, we cannot make a triangle with the side lengths of measurement 2 cm, 3 cm, and 10 cm. This is because sum of 2+3 < 10).

<h3>What is triangle inequality theorem?</h3>

Triangle inequality theorem of a triangle says that the sum of any of the two sides of a triangle is always greater than the third side.

Suppose a, b and c are the three sides of a triangle. Thus according to this theorem,

$(a+b) > c\\(b+c) > a\\(c+a) > b$

Now, for this case, the sides given are:

a =2 cm,
b = 3 cm,
and c = 10 cm

But we see that:

a+ b = 5 cm which is < c which is of 10 cm.

Thus, these lengths don't satisfy the triangle inequality theorem, and therefore, cannot be sides of any triangle.

Learn more about triangle inequality theorem here:

brainly.com/question/342881

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Translate the description as an algebraic expression: the square of the ratio of 11 and k

damaskus [11]

Answer:

The given description is equivalent to the algebraic expression,

$(\frac {11}{k})^{2}$

Step-by-step explanation:

The given description is ,

'The square of the ratio of 11 and k'

which is equivalent to the algebraic expression,

$(\frac {11}{k})^{2}$

6 0

3 years ago

Triangle S and Triangle L are scaled copies of one

tatiyna

Answer:

Step-by-step explanation:

The bottom of S is 2 units and the bottom of L is 4 Units.

4:2 is the ratio, and that is equal to 2

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2 years ago

Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

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2 years ago

9. (3x + 4y)² – (3z) 2square

rodikova [14]

Solution: (3x + 4y - 3z) x (3x + 4y + 3z)

5 0

3 years ago

Read 2 more answers

What does it mean by type of angles and relationship between the angles if the line are parallel ?

valkas [14]

Where two lines cross, four (4) angles are formed. Two adjacent angles that share a vertex can be called a "linear pair." Two non-adjacent angles that have the same vertex and the same lines for sides, are called "vertical angles."

Where a transversal crosses two parallel lines, eight (8) angles are formed, four (4) at each intersection. In addition to the above angle types, there are other relationships that can be described.

Angles can be called corresponding, if they are in the same location relative to the intersection point (both on the north-east corner, for example). Angles are interior if they are between the parallel lines; exterior of they are not. They get the additional descriptor of opposite, if they are on opposite sides of the transversal.

If the angles are not opposite, and share a side but not a vertex, they can be called adjacent. (Only interior angles can be adjacent, one pair on either side of the transversal.)

_____

With the above definitions in mind, several relationships arise where a transversal crosses parallel lines. (Types of angles are italicized.) Two examples of each type are listed by angle numbers. Angles 1 and 2 are a linear pair, for example.

Angles of a linear pair are supplementary. (1, 2), (5, 7)

Vertical angles are congruent. (1, 4), (6, 7)

Adjacent angles are supplementary. (3, 5), (4, 6)

Corresponding angles are congruent. (1, 5), (4, 8)

Opposite interior angles are congruent. (3, 6), (4, 5)

Opposite exterior angles are congruent. (1, 8), (2, 7)

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3 years ago