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Bond [772]
4 years ago
10

-16 - 6v = -2 (8v -7)

Mathematics
1 answer:
sashaice [31]4 years ago
5 0

-16-6v = -16v+14

-6v= -16v+30

10v=30

v=3

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Find four distinct complex numbers (which are neither purely imaginary nor purely real) such that each has an absolute value of
Luda [366]

Answer:

  • 0.5 + 2.985i
  • 1 + 2.828i
  • 1.5 + 2.598i
  • 2 + 2.236i

Explanation:

Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.

Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.

The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:

  • |C| = \sqrt{a^2+b^2}

Then, the work consists in finding pairs (a,b) for which:

  • \sqrt{a^2+b^2}=3

You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}

I will use b =0.5, b = 1, b = 1.5, b = 2

b=0.5;a=\sqrt{9-0.5^2}=2.958\\ \\b=1;a=\sqrt{9-1^2}=2.828\\ \\b=1.5;a=\sqrt{9-1.5^2}=2.598\\ \\b=2;a=\sqrt{9-2^2}=2.236

Then, four distinct complex numbers that have an absolute value of 3 are:

  • 0.5 + 2.985i
  • 1 + 2.828i
  • 1.5 + 2.598i
  • 2 + 2.236i
4 0
4 years ago
In the box, complete the first 4 steps for graphing the quadratic function given. (Use ^ on the keyboard to indicate an exponent
Vaselesa [24]
Step 1: convert the equation into fhe vertex form

To do you can complete squares:

y = - [x^2 + 4x + 3]

y = - [ (x + 2)^2 - 4 + 3]

y = - [ (x + 2)^2 - 1] = - (x+2)^2 + 1

Then the vertex is (-2, 1)

Now you can drasw the vertex


Step 2: Find the roots (zeros)

y = - [ (x + 2)^2 - 1] = 0

(x + 2)^2 - 1 = 0

(x+2)^2 = 1

(x+2) = (+/-) √1

x + 2 = (+/-1)

x = - 2 +/- 1

x = -1 and x = -3

Now you draw the points (-1,0) , (-3,0)

Step 3: find the interception with the y-axis.

That is y value when x = 0

y = - (0)^2 - 4(0) - 3 = -3

Then draw the point (0, -3)

Step 4: given that the coefficent of x is negative (-1) the parabola is open downward.

So, with those four points: vertex (-2,1), (-1,0), (-3,0) and (0,-3), you can sketch the function.
7 0
3 years ago
(4,5);y=3/2x+3 write an equation of the line that passes through the given point that is parallel to the given line
Aleonysh [2.5K]

Answer:

2x-3/y-5/37=0

Step-by-step explanation:

here, given equation of line is

y=3/2x+3

or,y(2x+3)=3

or,2x+3=3/y

or,2x-3/y+3=0...eqn(i)

equation of any line parallel to line (i) is

2x-3/y+k=0.. eqn(ii)

since, the line passes through (4,5)[replacing x=4 and y=5 in eqn(ii), we get]

2*4-3/5+k=0

or,8-3/5+k=0

or,40-3/5+k=0

or,k=-5/37

substituting the value of k=-5/37 in eqn(ii)

2x-3/y-5/37=0 is the required equation of line.

3 0
3 years ago
PLEASE IF YOUR A MATH EXPERT ANSWER THIS I'M LOSING MY SANITY
Nana76 [90]

Answer:

2

Step-by-step explanation:

Plug in the values for x and y into the expression and simplify:

numerator: 4(2) + 4 = 12

denominator: 2(4) - 2 = 6

dividing: 12/6=2

8 0
3 years ago
Read 2 more answers
True or false? <br><br> this is the equation of a horizontal hyperbola (y-2)^2/16-(x+1)^2/144=1
Mars2501 [29]
<span>its false I hope I helped</span>
4 0
3 years ago
Read 2 more answers
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