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enyata [817]
3 years ago
11

Step by step how to get 22.04 times 2.8 with work

Mathematics
1 answer:
Bumek [7]3 years ago
4 0

Answer:

61.712

Step-by-step explanation:

Multiply .8x.04

Multiply .8x.0

Multiply .8x2

Multiply .8x2

Than add a zero under the answer of .8x.04

Multiply 2x.04

Multiply 2x.0

Multiply 2x2

Multiply 2x 2

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The tallest building in Africa is the Carlton Centre in Johannesburg, South Africa. What is the distance from the top of this 73
Contact [7]

Answer:

b = 33 miles

Step-by-step explanation:

The illustration forms a right angled triangle. The opposite side of the triangle is where we are looking for . The adjacent side is the radius 4000 miles.

5280 ft = 1 mile

735 ft = ? miles

cross multiply

735/5280

miles = 0.13920454545  miles

The hypotenuse side is the height of the building + 4000 miles = 0.140 + 4000 = 4000.14  miles

using Pythagoras theorem

c² = a² + b²

c = 4000.14

a = 4000

4000.14² = 4000² + b²

b² = 4000.14² - 4000²

b² = 16001120.0196  - 16000000

b² = 1120.0196

b = √1120.0196

b = 33.4666938911

b = 33 miles

3 0
3 years ago
GEOMETRY SIDE LENGTHS​
PilotLPTM [1.2K]

Answer:

I believe that the third answer is correct..... but I'm not positive.

Step-by-step explanation:

if you get the third side of the triangle to be 71 then it would go first. Then it would go to 45-55. And lastly to 71-45

3 0
3 years ago
What is the approximate volume of the sphere?
Vinvika [58]
C. 523.33

the formula is 4/3*3.14*r^3
R being radius (:
4 0
3 years ago
Read 2 more answers
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
What is the slope of this graph?
larisa [96]
Since the slope is falling, instead of rising, it would be negative. 

Now, on the line, I plotted (0, 2) and (-2, 10).

You would now plug this into the rise/run formula. 
The formula is \frac{y2 - y1}{x2 - x1}

The equation would become, 
\frac{10 - 2}{-2 - 0} = \frac{8}{-2} = -4

I believe the answer would be -4. 

I hope this helps!

6 0
3 years ago
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