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natita [175]
3 years ago
12

G(x) = 4x + 3; Find g(10)

Mathematics
2 answers:
adelina 88 [10]3 years ago
8 0

Answer:

43

Step-by-step explanation:

g(x)=4x+3

plug in 10

g(10)= 4(10)+3

      = 40+3

      = 43

natta225 [31]3 years ago
8 0

Answer:

Sorry I just needed points

Step-by-step explanation:

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Two people can swing on a trapeze at the same time as long as
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Yes!

Step-by-step explanation:

Joe’s weight with Jake’s weight is 490, which is less than 500, so yes!

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Find the following for the function f(x) = 3x² + 4x - 4.
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C_3fihcahjzshshsushshshxhxh

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jacobs puzzle has 84 pieces. Jacobs puts together 27 pieces in the morning. he puts together 38 more pieces in the afternoon. ho
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Find the amount of cardboard needed to make a box for a single slice of pizza. The box is shoen in the shape of a triangular pri
Molodets [167]
For this case, what you should do is calculate the surface area of the figure.
 We have then:
 Triangles:
 A1 = (1/2) * (7) * (12)
 A1 = 42 in ^ 2
 Rectangles:
 A2 = (1) * (12.5)
 A2 = 12.5 in ^ 2
 A3 = (1) * (7)
 A3 = 7 in ^ 2
 Finally, the total surface area is:
 A = 2A1 + 2A2 + A3
 Substituting values:
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 A = 84 + 25 + 7
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 Answer:
 
the amount of cardboard needed to make a box for a single slice of pizza is:
 
A = 116 in ^ 2
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
3 years ago
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