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katrin2010 [14]
3 years ago
15

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<br> prism shown by the net?
brilliants [131]

Answer:

80

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:

  1. 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.
  2. <em>Then for the x direction</em> 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}
  3. In the same fashion, <em>for the y direction</em>, 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}
8 0
2 years ago
Two years ago, Paul bought $350 worth of stock in a cell phone company. Since then the value of his stock has been increasing at
zalisa [80]

Answer: $384.13 ( rounded)

Original number: $384.125

Step-by-step explanation:

Find 9.75% of Paul's original stock, to find out how much it increased.

9.75*350/100=34.125

Add $34.125 to his original stock.

$384.125

5 0
3 years ago
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