Find the quotient of 90 over -10

2 answers:

5 0

Hi! The answer is -9 because 90/-10 = -9
I hope this helps you, Goodluck! :)

3 0

90/-10

= 9/-1

= -9

So, -9 is the quotient.

You might be interested in

mel-nik [20]

Using derivatives, it is found that the best estimate of f '(2) based on this table of values is of 10.

The rate of change from x = 0 to x = 2 is given by:

$r_1 = \frac{2 - (-16)}{2 - 0} = \frac{18}{2} = 9$

From x = 2 to x = 4, it is given by:

$r_2 = \frac{24 - 2}{4 - 2} = \frac{22}{2} = 11$

The average of these rates is:

$A = \frac{r_1 + r_2}{2} = \frac{9 + 11}{2} = 10$

Hence, the best estimate of f '(2) based on this table of values is of 10.

To learn more about derivatives, brainly.com/question/18590720

6 0

1 year ago

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
Then for the x direction it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, for the y direction, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$