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UNO [17]
3 years ago
5

What is the absolute value of the following complex number? -6+3i

Mathematics
2 answers:
max2010maxim [7]3 years ago
8 0

Answer:

Step-by-step explanation:

|-6+3i|=√((-6)²+(3)²)=√45=3√5

juin [17]3 years ago
5 0
6 and 3

All you gotta do count how many values it is away for 0
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Can someone please answer this for me I'm begging
GrogVix [38]
If the length of JK is 16.4 miles, it will be diameter of this circle, the radius will be JK/2 = 8.2 miles

As we know the length of circle is 2π*r  in this case r = 8.2

The arc JM is 1/4 length of the whole circle so arc JM = (2π*8.2)*(1/4) = 2π*8.2/4 = π*8.2/2 = π*4.1

if we assume that π=3.14 the answer will be 3.14 * 4.1 = 12.874.... And as the answer needs to be rounded to the nearest tenth it will be 12.874≈12.9

So the answer will be 12.9
8 0
3 years ago
Find the equation of the straight line<br> that passes through (5, 6) and (5, −6).
GalinKa [24]

9514 1404 393

Answer:

  x = 5

Step-by-step explanation:

These points have the same x-coordinate, so are on the same vertical line:

  x = 5

6 0
2 years ago
During the final 5 seconds of a race a cyclist increased her velocity from 4 m/s to 7 m/s . What was her average acceleration du
GREYUIT [131]

Answer:

The average acceleration of the cyclist was 0.6 m/s².

Step-by-step explanation:

Acceleration:

The rate change of velocity per unit time is call the acceleration of the object.

a=\frac{v-u}t

u= initial velocity

v= final velocity

t= time taken change of velocity.

During the final 5 seconds of a race cyclist increased her velocity from 4m/s to 7 m/s

Here v= 7 m/s and u=4 m/s t=5 seconds

\therefore a=\frac{7\ m/s-4 \ m/s}{5s}

     =\frac{3}{5} \ m/s^2

     =0.6 m/s²

The average acceleration of the cyclist was 0.6 m/s².

8 0
3 years ago
I need some help please
Veronika [31]
The definition of the tangent function tells you
   tan(angle) = (300 ft) / (distance to mountain)
This equation can be rearranged to
   (distance to mountain) = (300 ft) / tan(angle)

For the far end of the river,
   distance to far end = (300 ft) / tan(24°) ≈ 673.8 ft

For the near end of the river
   distance to near end = (300 ft) / tan(40°) ≈ 357.5 ft

Then the width of the river can be calculated by finding the difference of these distances:
   width of river = distance to far end - distance to near end
   width of river = 673.8 ft - 357.5 ft
   width of river = 316.3 ft

The appropriate answer choice is
   316 ft.
8 0
3 years ago
If G is the midpoint of FH, FG= 14x + 25 and GH = 73- 2x, find FH
Alenkinab [10]
Given that G is the midpoint of FH, then:
FG=GH
hence;
14x+25=73-2x
solving for x we get:
14x+2x=73-25
16x=48
x=48/16
x=3
therefore the length FG=14x+25 will be:
14*3+25
=67
hence FH will be:
FH=67*2=134
4 0
3 years ago
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