What is the absolute value of 9.8 and

Mathematics

1 answer:

Citrus2011 [14]3 years ago

7 0

The answer is -10/3 your answer

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the initial vertical velocity of a rocket shot straight up is 42 meters per second. How long does it take for the rocket to retu

Snowcat [4.5K]

It takes 4.3 seconds for the rocket to return to earth.

The equation is:
$0=-9.8t^2+v_0t+h_0$
where -9.8m/sec² is the acceleration due to gravity, v₀ is the initial velocity, and h₀ is the initial height. We will go from the assumption that the rocket is launched from the ground, so h₀=0, and we are told that the initial velocity, v₀, is 42. This gives us:

$0=-9.8t^2+42t$

We will use the quadratic formula to solve this. The quadratic formula is:
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Plugging in our information we have:
$t=\frac{-42\pm \sqrt{42^2-4(-9.8)(0)}}{2(-9.8)}
\\
\\=\frac{-42\pm \sqrt{1764-0}}{-19.6}
\\
\\=\frac{-42\pm \sqrt{1764}}{-19.6}
\\
\\=\frac{-42\pm 42}{-19.6}=\frac{-42-42}{-19.6} \text{ or } \frac{-42+42}{-19.6}
\\
\\=\frac{-84}{-19.6}\text{ or }\frac{0}{-19.6}
\\
\\=4.3\text{ or }0$

x=0 is when the rocket is launched; x=4.3 is when the rocket lands.

8 0

3 years ago

Can someone please help thanks

USPshnik [31]

Ohh this is easy ok so count the numbers and all of them are the number part like 16-32 16+16=32 keep going on like that for both problem and explain how u did it so ur teacher knows you get it now

8 0

3 years ago

Read 2 more answers

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$