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2 years ago

5

What is the value of the expression given below? (8-3i)-(8-3i)(8+8i)

2 answers:

enyata [817]2 years ago

6 0

Answer:

-76-43i

Step-by-step explanation:

First expand the multiplied terms

8-3i-(64+64i-24i+24)

8-3i-(64+40i+24)

Simplify

8-3i-64-40i-24

-76-43i

Lapatulllka [165]2 years ago

5 0

Answer:

$-80-43i$

Step-by-step explanation:

$(8-3i)-(8-3i)(8+8i)$

To find the value of the expression we need to simplify it

LEts multiply the parenthesis part using FOIL method

$(8-3i)-(64+64i-24i-24i^2)$

$(8-3i)-(64+40i-24i^2)$

The value of i^2 is -1

$(8-3i)-(64+40i-24(-1))$

$(8-3i)-(64+40i+24)$

$(8-3i)-(88+40i)$

Now we remove the parenthesis

$8-3i-88-40i$

combine like terms

$-80-43i$

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2 years ago

A baseball player got 102 hits in the last 300 times at bat. Explain how you would find the percent of times at bat.

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3 years ago

Read 2 more answers

Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.

FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

$\rho = \sqrt{x^{2}+y^{2}+z^{2}}$

$\phi = cos^{-1}\frac{z}{\rho}$

For angle θ:

If x > 0 and y > 0: $\theta = tan^{-1}\frac{y}{x}$ ;

If x < 0: $\theta = \pi + tan^{-1}\frac{y}{x}$ ;
If x > 0 and y < 0: $\theta = 2\pi + tan^{-1}\frac{y}{x}$ ;

Calculating:

a) (4,2,-4)

$\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}$ = 6

$\phi = cos^{-1}(\frac{-4}{6})$

$\phi = cos^{-1}(\frac{-2}{3})$

For θ, choose 1st option:

$\theta = tan^{-1}(\frac{2}{4})$

$\theta = tan^{-1}(\frac{1}{2})$

b) (0,8,15)

$\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}$ = 17

$\phi = cos^{-1}(\frac{15}{17})$

$\theta = tan^{-1}\frac{y}{x}$

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = $\frac{\pi}{2}$

c) (√2,1,1)

$\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}$ = 2

$\phi = cos^{-1}(\frac{1}{2})$

$\phi$ = $\frac{\pi}{3}$

$\theta = tan^{-1}\frac{1}{\sqrt{2} }$

d) (−2√3,−2,3)

$\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}$ = 5

$\phi = cos^{-1}(\frac{3}{5})$

Since x < 0, use 2nd option:

$\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }$

$\theta = \pi + \frac{\pi}{6}$

$\theta = \frac{7\pi}{6}$

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

$r=\sqrt{x^{2}+y^{2}}$

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

$r=\sqrt{4^{2}+2^{2}}$ = $\sqrt{20}$

$\theta = tan^{-1}\frac{1}{2}$

z = -4

b) (0, 8, 15)

$r=\sqrt{0^{2}+8^{2}}$ = 8

$\theta = \frac{\pi}{2}$

z = 15

c) (√2,1,1)

$r=\sqrt{(\sqrt{2} )^{2}+1^{2}}$ = $\sqrt{3}$

$\theta = \frac{\pi}{3}$

z = 1

d) (−2√3,−2,3)

$r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}$ = 4

$\theta = \frac{7\pi}{6}$

z = 3

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3 years ago