7.5 = x + 1.25 please answer

Mathematics

1 answer:

Levart [38]3 years ago

5 0

Move +1.25 to the left side of the equal sign:

7.5-1.25=x

subtract 7.5 and 1.25 which is...

6.25=x

You might be interested in

Write the complex number in polar form:

emmasim [6.3K]

9514 1404 393

Answer:

2√30 ∠-120°

Step-by-step explanation:

The modulus is ...

√((-√30)² +(-3√10)²) = √(30 +90) = √120 = 2√30

The argument is ...

arctan(-3√10/-√30) = arctan(√3) = -120° . . . . a 3rd-quadrant angle

The polar form of the number can be written as ...

(2√30)∠-120°

_____

Additional comments

Any of a number of other formats can be used, including ...

(2√30)cis(-120°)

(2√30; -120°)

(2√30; -2π/3)

2√30·e^(i4π/3)

Of course, the angle -120° (-2π/3 radians) is the same as 240° (4π/3 radians).

At least one app I use differentiates between (x, y) and (r; θ) by the use of a semicolon to separate the modulus and argument of polar form coordinates. I find that useful, as a pair of numbers (10.95, 4.19) by itself does not convey the fact that it represents polar coordinates. As you may have guessed, my personal preference is for the notation 10.95∠4.19. (The lack of a ° symbol indicates the angle is in radians.)

6 0

3 years ago

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

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: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$