Help meee i need the X and what the angles are

Mathematics

1 answer:

algol133 years ago

6 0

Answer:

x will be 120 degrees

Step-by-step explanation:

I will loves easy questions!

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Can anyone help me please

Sauron [17]

Range of values of x is 25°<x<27°.

<u>Step-by-step explanation:</u>

An angle is formed when two rays, a line with one endpoint, meet at one point called a vertex.Alternate interior angles are formed when a transversal passes through two lines. The angles that are formed on opposite sides of the transversal and inside the two lines are alternate interior angles. The Alternate Interior Angles theorem states, if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.

In above question, sides are equal and so alternate interior angles are equal i.e. 2x + 10° = 62° ⇒ 2x = 52° ⇒ x =26°

∴ range of values of x is 25°<x<27°

4 0

4 years ago

It takes Ebru 131313 minutes to bike to school. If she walks, it takes her twice as long. Ebru leaves for school at 7{:}157:157,

sdas [7]

Answer: 7:41 am

Step-by-step explanation:

It takes Ebru 13 minutes to bike to school but if she walks it would take her twice at long.

= 13 * 2

= 26 minutes

If she leaves the house at 7:15 , she will arrive at:

= 715 + 26

= 7:41 hours

3 0

2 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is