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emmainna [20.7K]
3 years ago
8

How many times larger is the value of 42.38 compared to 4.238

Mathematics
1 answer:
patriot [66]3 years ago
5 0

Answer:

10 times larger

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The length of a football playing field is 100 yards between the opposing goal lines. What is the length of the football field in
sweet-ann [11.9K]

Answer:

33.3

Step-by-step explanation:

100 divided by 3

3 0
3 years ago
Which descriptions can describe more than one triangle? check all that apply
babunello [35]

Answer:

Angle measurements of 35°, 35°, and 110°

Angle measurements of 40°, 60°, and 80°

Step-by-step explanation:

If you specify only the three angles, you can have an infinite number of similar triangles of different sizes.

A is wrong. There can be only one triangle in which all three side lengths are specified.

C is wrong. The angles do not add up to 180°.

5 0
3 years ago
Read 2 more answers
-4 3/5 divided by 1 1/5 as a mixed number or simplified
WITCHER [35]
-23 over 5 ÷ 11 over 5
-23 × 5 over 5 × 5
-115 over 55


Therefore, your answer would be: -2 and 1 over 11
4 0
3 years ago
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How to figure out median weight
statuscvo [17]
They need look a median wieght.
7 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
3 years ago
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