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3 years ago

It took14 min to walk around the block 4times. What was her unit rate?

Mathematics

2 answers:

Harman [31]3 years ago

6 0

The unit rate is 3.5.

Luba_88 [7]3 years ago

3 0

7.5 min per walk around the block

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Assume z = x + iy, then find a complex number z satisfying the given equation. d. 2z8 – 2z4 + 1 = 0

kodGreya [7K]

Answer: complex equations has n number of solutions, been n the equation degree. In this case:

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}$

Step-by-step explanation:

I start with a variable substitution:

$Z^{4} = X$

Then:

$2X^{2}-2X+1=0$

Solving the quadratic equation:

$X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}$

$X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.$

Replacing for the original variable:

$Z=\sqrt[4]{0,5+0,5i}$

or $Z=\sqrt[4]{0,5-0,5i}$

Remembering that complex numbers can be written as:

$Z=a+ib=|Z|e^{ic}$

Using this:

$Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.$

Solving for the modulus and the angle:

$Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.$

The possible angle respond to:

$RAng_{12...n} =\frac{Ang +360*(i-1)}{n}$

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"

In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º

Obtaining 8 different angles, therefore 8 different solutions.

3 0

3 years ago

Assume that f(x)=ln(1+x) is the given function and that Pn represents the nth Taylor Polynomial centered at x=0. Find the least

WINSTONCH [101]

Answer:

the least integer for n is 2

Step-by-step explanation:

We are given;

f(x) = ln(1+x)

centered at x=0

Pn(0.2)

Error < 0.01

We will use the format;

[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01

So;

f(x) = ln(1+x)

First derivative: f'(x) = 1/(x + 1) < 0! = 1

2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1

3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2

4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6

This follows that;

Max|f^(n+1) (c)| < n!

Thus, error is;

(n!/(n + 1)!) × 0.2^(n + 1) < 0.01

This gives;

(1/(n + 1)) × 0.2^(n + 1) < 0.01

Let's try n = 1

(1/(1 + 1)) × 0.2^(1 + 1) = 0.02

This is greater than 0.01 and so it will not work.

Let's try n = 2

(1/(2 + 1)) × 0.2^(2 + 1) = 0.00267

This is less than 0.01.

So,the least integer for n is 2

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4 years ago

You order a pizza for dinner

ch4aika [34]

The radius is half the diameter. So the diameter is 14.

7 0

3 years ago

Solve the equation 3y-9x=-3

ipn [44]

This equation can have a number of solutions. Typically when solving a two-variable equation, you are looking for the slope intercept form of the line. In this case that is:

y = 3x - 1

7 0

3 years ago

Solve for x. -3+85-5=-8 A. -2. B.-1 C.-3/4 D.0

posledela

Answer:

-2

Step-by-step explanation:

-3+8x-5=-8

8x = -16

x = -2

5 0

3 years ago

Read 2 more answers