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3 years ago

How much larger is 6 x 105 compared to 2 x 103? Explain how you know.

Mathematics

2 answers:

Lorico [155]3 years ago

4 0

Answer:

397

Step-by-step explanation:

well we know that 6*105= 630 and 2*103 = 206 so we subtract 206 from 630 giving us 397

hope this helps.

34kurt3 years ago

4 0

Answer:

6×105=630

Step-by-step explanation:

2×103=206 so 6×105 is greater with 424 yimes2×103.

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6 = 1 – 2n + 5 The answers are n=0 n=6 No solution All real numbers

svp [43]

Answer:

The answer is n=0

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3 years ago

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Z^4-5(1+2i)z^2+24-10i=0

mixer [17]

Using the quadratic formula, we solve for $z^2$ .

$z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2$

Taking square roots on both sides, we end up with

$z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}$

Compute the square roots of -171 + 140i.

$|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221$

$\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)$

By de Moivre's theorem,

$\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i$

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

$t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}$

as well as the fact that

$0$

$\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}$

(whose signs are positive because of the domain of $\frac t2$ ).

This leaves us with

$z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}$

Compute the square roots of 5 + 12i.

$|5 + 12i| = \sqrt{5^2 + 12^2} = 13$

$\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)$

By de Moivre,

$\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i$

and its negative, -3 - 2i. We use similar reasoning as before:

$t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}$

$1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4$

$\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}$

Lastly, compute the roots of -2i.

$|-2i| = 2$

$\arg(-2i) = -\dfrac\pi2$

$\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i$

as well as -1 + i.

So our simplified solutions to the quartic are

$\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}$

3 0

1 year ago

How do you solve? 2(1+4n)+10(n-7)=2n-n

stiv31 [10]

2(1+4n)+10(n-7)=2n-n
distribute
2+8n+10n-70=2n-n
combine like terms
18n-68=n
move n to other side
17n=68
divide
n=4

4 0

3 years ago

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

REY [17]

Answer:

the first number is 3.4 and the second number is 1

Step-by-step explanation:

The computation of the two numbers is shown below:

Let us assume the first number be x

And, the second number be y

Now the equations are

5x + 4y = 21

And,

5x + 9y = 16

Now subtract equation 2 from equation 1

-5y = -5

y = 1

And, the x is

5x + 4 = 21

5x = 17

x = 3.4

Hence, the first number is 3.4 and the second number is 1

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3 years ago

A circular plate has a crack along the line AB, as shown below: A circular plate is shown. The center of the plate is labeled as

makkiz [27]

Well, since point A represents the center, and B represents a point on the outer line of the circle, segment AB would represent the radius, since the radius represents the length from the center to the outside of a circle

Hope this helps

8 0

4 years ago

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