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3 years ago

True or False? The segment AB is congruent to the segment BC. A. True B. False

Mathematics

1 answer:

AleksAgata [21]3 years ago

3 0

Answer:

True

Step-by-step explanation:

If O to the edge of the circle is the radius and if B is on the radius line and has a right angle, then B is the halfway point.

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Prime factors of 20 i dont know this question

dem82 [27]

Answer:

2 x 2 x 5,

Step-by-step explanation:

20 = 1 x 20

2 x 10

4 x 5

Factors of 20= 1, 2, 4, 5, 10, 20.

Prime factors are = 2 x 2 x 5,

6 0

4 years ago

Read 2 more answers

Given f(x) = x3 – 2x2 – x + 2, the roots of f(x) are

kolezko [41]

You have the following expression given in the problem:

f(x) = x³ – 2x²<span> – x + 2

Therefore, to find the roots, you must apply the proccedure shown below:

1. You have:

0 </span><span>= x3 – 2x2 – x + 2

2. Then, when you factor, you obtain:

(x-2)(x-1)(x+1)=0

3. Therefore, you have that the roots are the following:

x1=-1
x2=1
x3=2</span>

3 0

3 years ago

Read 2 more answers

The length of a rectangle is 1 foot more than twice the width. The area is 55 square feet. Find the length of the dimension of t

Dvinal [7]

Answer:

11 ft

Step-by-step explanation:

The formula for the area of a rectangle of length L and width W is

A = L * W.

Represent the length, L, by 2W + 1 ("1 more than twice the width"). Then the area formula (above) becomes:

A = 55 ft^2 = (2W + 1)(W), which, when simplified, results in:

2W^2 + W - 55 = 0, whose coefficients are {2, 1, -55}. Let's use the quadratic formula to solve this for W:

The discriminant, b^2 - 4ac, is 1^2 - 4(2)(-55), or 441. The square root of 441 is 21. Thus, the quadratic formula gives us:

-1 ± 21

W = ------------- = (1/4)(-1 + 21), or W = 5 and W = -22/4 = -11/2

2(2)

A measure of length can't be negative, so we reject W = -11/2 in favor of W = 5.

Then, according to the formula for L that we found earlier, L = 2W + 1, or

L = 2(5) + 1, or L = 11.

Check: Does L * W result in an area of 55 ft²? Yes

The length dimension of the rectangle is 11 ft

3 0

3 years ago

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,

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