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yuradex [85]
3 years ago
9

What is the value of the product (3 – 2i)(3 + 2i)? 5 9 + 4i 9 – 4i 13

Mathematics
2 answers:
Assoli18 [71]3 years ago
8 0

Answer:

13

Step-by-step explanation:

(3-2i)(3+2i)

as (a+b)(a-b) =a²-b²

3²-(2i)²

=9-(4*-1)

=9-(-4)

=9+4

=13

Hope this helps you!

goldfiish [28.3K]3 years ago
6 0

Answer:

13

Step-by-step explanation:

3^2=9, 2i*(-2i)=-4i^2=4

the 6i and -6i cancel leaving the positive 9 and 4

9+4=13

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When a cube with sides numbered one through six is roll one time what is the probability of rolling a four or greater?
Elanso [62]
Since it's a six dice then it's out of x/6

How much is 4 to 6 ?

~3 (counting 4) 4.5.6...

So it's 3/6 = 1/2

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3 years ago
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What is the solution for this inequality? 3(x-4)>5(x-2)-8
Marizza181 [45]

3x - 12 > 5x - 10 - 8 \\ 10 + 8 - 12 > 5x - 3x \\ 6 > 2x \\ x < 3
7 0
3 years ago
the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function
Likurg_2 [28]

The function of the area of the square is A(t)=121t^{2}

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

\frac{dx}{dt} = 11 \frac{cm}{sec}

⇒x=11t

Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about The function of the area of the square is A(t)=121t^{2}

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

\frac{dx}{dt} = 11 \frac{cm}{sec}

⇒x=11t

Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about area here:

brainly.com/question/27683633

#SPJ4

3 0
2 years ago
Find the length of the missing side. triangle with an 8 inch side and 12 inch side with a right angle 8.9 in. 104 in. 4 in 14.4
Mamont248 [21]

Given:

In a triangle, length of one side is 8 inches and length of another side is 12 inches, and an angle is a right angle.

To find:

The length of the missing side.

Solution:

In a right angle triangle,

Hypotenuse^2=Perpendicular^2+Base^2

Suppose the measures of sides adjacent to the right angle are 8 inches and 12 inches.  

Substituting Perpendicular = 8 inches and Base = 12 inches, we get

Hypotenuse^2=8^2+12^2

Hypotenuse^2=64+144

Hypotenuse^2=208

Taking square root on both sides, we get

Hypotenuse=\sqrt{208}

Hypotenuse=14.422205

Hypotenuse\approx 14.4

The length of the missing side is 14.4 inches. Therefore, the correct option is D.

4 0
2 years ago
X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square
swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = -\frac{4}{5}x+\frac{58}{5}

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

   [x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

   (x + 2)² + (y - 5)²- 29 + 20 = 30

   (x + 2)² + (y - 5)²- 9 = 30

   (x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

    m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \frac{10-5}{2+2}

          = \frac{5}{4}

    Let the slope of the tangent which is perpendicular to this line is 'm_{2}'

    Then by the property of perpendicular lines,

          m_{1}\times m_{2}=-1

          \frac{5}{4}\times m_{2}=-1

                 m_{2}=-\frac{4}{5}

   Now the equation of the line passing though (2, 10) having slope m_{2}=-\frac{4}{5}

           y - y' = m_{2}(x-x')

           y - 10 = -\frac{4}{5}(x-2)

           y - 10 = -\frac{4}{5}x+\frac{8}{5}

                  y = -\frac{4}{5}x+\frac{8}{5}+10

                  y = -\frac{4}{5}x+\frac{58}{5}

Therefore, equation of the line will be, y = -\frac{4}{5}x+\frac{58}{5}

7 0
3 years ago
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