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3 years ago

Pleasr Help will mark brainly...! How are actual rocket launches similar to how balloon rockets and canister rockets launch?

Mathematics

2 answers:

Sever21 [200]3 years ago

7 0

Answer:

Publicado por primera vez el 13 de octubre de 2006; revisión sustantiva Lun 17 de julio de 2017

Un ciudadano es un miembro de una comunidad política que disfruta de los derechos y asume los deberes de membresía. Esta amplia definición se percibe, con pequeñas variaciones, en las obras de autores contemporáneos, así como en la entrada "citoyen" de la Encyclopédie [1753] de Diderot y d'Alembert [1]. A pesar de este punto de partida común y de algunas referencias compartidas, [2] las diferencias entre las discusiones del siglo XVIII y los debates contemporáneos son significativas.

explanation:

espero que ayude.

Vika [28.1K]3 years ago

5 0

Answer:

What to do

Tie one end of the string to a chair, door knob, or other support.

Put the other end of the string through the straw.

Pull the string tight and tie it to another support in the room.

Blow up the balloon (but don't tie it.) ...

Let go and watch the rocket fly!

Love, gocry

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Step-by-step explanation:

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4 years ago

Trigonometric functions: what is the value of x?

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x ≈ 19.7

Step-by-step explanation:

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3 years ago

If sin theta = 8/17 and cot theta < 0, what is sec theta?

klasskru [66]

Answer:

$-\frac{17}{15}$

Step-by-step explanation:

By definition, $\cot \theta=\frac{1}{\tan \theta}$ and $\sec \theta=\frac{1}{\cos \theta}$ . Since since $\cot \theta$ is negative, $\tan \theta$ must also be negative, and since $\sin \theta$ is positive, we must be in Quadrant II.

In a right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. The cosine of an angle in a right triangle is equal to its adjacent side divided by the hypotenuse. Therefore, we can draw a right triangle in Quadrant II, where the opposite side to angle theta is 8 and the hypotenuse of the triangle is 17.

To find the remaining leg, use to the Pythagorean Theorem, where $a^2+b^2=c^2$ , where $c$ is the hypotenuse, or longest side, of the right triangle and $a$ and $b$ are the two legs of the right triangle.

Solving, we get:

$8^2+b^2=17^2,\\b^2=17^2-8^2,\\b^2=\sqrt{17^2-8^2}=\sqrt{225}=15$

Since all values of cosine theta are negative in Quadrant II, all values of secant theta must also be negative in Quadrant II.

Thus, we have:

$\sec\theta=\frac{1}{\cos \theta}=-\frac{1}{\frac{15}{17}}=\boxed{-\frac{17}{15}}$

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3 years ago