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3 years ago

3. Suppose m = 2 + 6i, and |m+n|=3√10, where n is a complex number.

Mathematics

1 answer:

Anestetic [448]3 years ago

7 0

Answer:

Step-by-step explanation:

Suppose m = 2 + 6i, and |m+n|=3√10,

The modulus sign means m+n can either be positive or negative as shown.

If it is positive:.2+6i+n = 3√10

n = 3√10-(2+6i)

n = 3√10-2-18√10i

n = (-2+3√10)+√10i

b) Example of the complex number is given as (-2+3√10)+√10i. This is a complex number because it contains the real part and the imaginary part.

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natima [27]

Answer:

I'm not 100% sure but I believe the answer is 27

because -48/-4=12

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3 years ago

Read 2 more answers

Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0

devlian [24]

Answer:

The root of the equation $x^3-0.2589x^{2}+0.02262x-0.001122=0$ is x ≈ 0.162035

Step-by-step explanation:

To find the roots of the equation $x^3-0.2589x^{2}+0.02262x-0.001122=0$ you can use the Newton-Raphson method.

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess $x_{0}$ for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

$x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}$

For the information given:

$f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262$

For the initial value $x_{0}$ you can choose $x_{0}=0$ although you can choose any value that you want.

So for approximation $x_{1}$

$x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021$

Next, with $x_{1}=0.0496021$ you put it into the equation

$f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150$ , you can see that this value is close to 0 but we need to refine our solution.

For approximation $x_{2}$

$x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883$

Again we put $x_{2}=0.168883$ into the equation

$f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307$ this value is close to 0 but again we need to refine our solution.

We can summarize this process in the following table.

The approximation $x_{5}$ gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.

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3 years ago

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Taya2010 [7]

C. polygons!!

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%2013%2B14%5Ctimes%2011" id="TexFormula1" title="\sf 13+14\times 11" alt="\sf 13+14\times

nlexa [21]

Answer:

167

Step-by-step explanation:

By BODMAS rule,

13 + 14 x 11

= 13 + ( 14 x 11 )

= 13 + 154

= 167

Note : -

B - Brackets

O - Of

D - Division

M - Multiplication

A - Addition

S - Subtraction

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Rachel budgeted $76.50 for school clothes. This is 0.45 of her total budget. How much does Rachel have in her total budget.

HACTEHA [7]

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