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Anarel [89]
3 years ago
13

PLEASE HELP!! WILL MARK BRAINLEST!!

Mathematics
2 answers:
Arturiano [62]3 years ago
8 0

Answer:

2.178 × 10^6

Step-by-step explanation:

Multiply 4840 × 450 to get 2,178,000

Move the decimal 6 places left then multiply by 10 to the 6th power, the number of places the decimal moved, to put it into scientific notation form. Scientific notation always has only one digit to the left of the decimal point.

alex41 [277]3 years ago
4 0
The answer is b 2.178 x 10 to the 6th power.
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Banana split has 7.857 calories
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Ira Lisetskai [31]

y increases as x increases

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What is the sum of ∞Σn=1 2(1/5)^n-1? s=1/5 s=2/5 s=5/3 s=5/2
lutik1710 [3]

The sum of the sum notation ∞Σn=1 2(1/5)^n-1 is S= 5/2

<h3>How to determine the sum of the notation?</h3>

The sum notation is given as:

∞Σn=1 2(1/5)^n-1

The above notation is a geometric sequence with the following parameters

  • Initial value, a = 2
  • Common ratio, r = 1/5

The sum is then calculated as

S = a/(1 - r)

The equation becomes

S = 2/(1 - 1/5)

Evaluate the difference

S = 2/(4/5)

Express the equation as products

S = 2 * 5/4

Solve the expression

S= 5/2

Hence, the sum of the sum notation ∞Σn=1 2(1/5)^n-1 is S= 5/2

Read more about sum notation at

brainly.com/question/542712

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7 0
2 years ago
11) The difference of two numbers is 3. Their sum is 23. What are the numbers?
Artyom0805 [142]

Answer:

Step-by-step explanation:

ill try this one in a few minutes for you.

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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