Describe the dimensions of a triangular prism that has a surface area between 550 and 700 square inches?

Mathematics

1 answer:

lina2011 [118]3 years ago

7 0

This is really hard question you need more description

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Please help and explain if you can

finlep [7]

You're pretty much screwed

4 0

3 years ago

Describe and correct the error a student made when graphing y+5=-3/4(x-8).

Elis [28]

In step 1 the y-intercept should be plotted at (0,-6)
Step-by-step explanation:
Remember that to find the Y intercept in any linear equation you need to use 0 as your X value, this means taking the formula in the y=mx+b form and replacing X with a 0.
Since the formula is y = -3/4x -6
We just insert a 0 insted of the "x"
y = -3/4(0) -6
y=0-6
y=6
So the y-intercept sould be placed in (0,-6)
That's what he did wrong when graphing the equation.

8 0

2 years ago

What is the surface area of the rectangular prism shown by the net?

brilliants [131]

Answer:

Step-by-step explanation:

There are four 4 by 3 rectangles, so I did 4x3 and got 12, then multiplied by 4 to get 48. There are also two 4 by 4 squares, so 4x4=16 and 16x2=32

32+48=80

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