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3 years ago

X^2 - 10x + 13 = 0 Solve quadratic equation In simplest form

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

$5 \pm 2\sqrt{3}$

Step-by-step explanation:

A quadratic equation in the form

$ax^2+bx+c=0$

Can be solved by using the formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In this problem, the equation is

$x^2-10x+13=0$

So, we have

a = 1

b = -10

c = +13

Using the equation above, we find:

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(13)}}{2(1)}=\frac{10 \pm \sqrt{48}}{2}=\frac{10 \pm 4\sqrt{3}}{2}= 5 \pm 2\sqrt{3}$

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3 years ago

what is the value of x in the diagram below? if necessary round your answer to the nearest tenth of a until

VladimirAG [237]

Answer:

Step-by-step explanation:

Calculate AC using Pythagoras' identity in ΔABC

AC² = 20² - 12² = 400 - 144 = 256, hence

AC = $\sqrt{256}$ = 16

Now find AD² from ΔACD and ΔABD

ΔACD → AD² = 16² - (20 - x)² = 256 - 400 + 40x - x²

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Equate both equations for AD², hence

256 - 400 + 40x - x² = 144 - x²

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3 years ago

Read 2 more answers

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

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3 years ago

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kvv77 [185]

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3 years ago

6. Find LCM (10,14,63).

Katen [24]

Answer:

630 is the LCM

Step-by-step explanation:

You have to list all the multiples of each number till you find the lowest same value one which in this case it's 630.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650

Multiples of 14:

14, 28, 42, 56, 70, 84, 98, 112, 126, 140, 154, 168, 182, 196, 210, 224, 238, 252, 266, 280, 294, 308, 322, 336, 350, 364, 378, 392, 406, 420, 434, 448, 462, 476, 490, 504, 518, 532, 546, 560, 574, 588, 602, 616, 630, 644, 658

Multiples of 63:

63, 126, 189, 252, 315, 378, 441, 504, 567, 630, 693, 756

3 0

3 years ago