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2 years ago

Find the value of 2 numbers if their sum is 9 and their difference is 1

Mathematics

2 answers:

Sloan [31]2 years ago

5 0

Makes no sense.

p.s i am a fourth grader

Kazeer [188]2 years ago

5 0

The number are 4 and 5.

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Which expression is equivalent to 10i/2-i a. –5i – 2 b. –4i – 2 c. 4i – 2 d. 5i – 2

scoray [572]

The answer to the question is
C. 4i-2

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3 years ago

Read 2 more answers

How to write six billion sixty- seven million four hundred four thousand thirteen and nineteen thousandths in standard form?

Neko [114]

Standard for would be 6,670,404,013.19

8 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Can you please help formate this problem

vovangra [49]

Take another picture

5 0

3 years ago

A blind man wanders under a door into an empty room. After running into the walls repeatedly for a while, it stops to think and

Nostrana [21]

From the side where the man is located in the 10' by 12' rectangular room, we have;

A) 0.046

B) The average time to escape in first attempt is approximately 605.02 seconds

<h3>Which concept can be used to find the odds and time of escape?</h3>

A) From The given diagram, we have;

Width of the door = 10 - 6 - 2 = 2

Therefore, the door is 2 feet wide

Distance, d1, from the man to the side of the door closer to him is given by Pythagorean theorem as follows;

d1 = √(6² + 7²) = √(85)

From the furthest side of the door to the man, we have;

d2 = √(7² + 8²) = √(113)

According to the law of cosines, we have;

2² = 85+113 - 2×√(85)×√(113) cos(A)

Where angle A is the angle formed by the lines from the man to the closer and farther sides of the door.

2×√(85)×√(113) cos(A) = 85+113 - 2²

A = arccos((85+113 - 2²)/(2×√(85)×√(113)))

A = arccos (194/(2×√(9605)) ≈ 8.213°

The angle of the possible directions in which the man can turn is 180°.

Therefore;

The probability, P(E), of escaping in the first move is therefore;

P(E) = 8.213°/180° ≈ 0.046

B) Given that the man escapes in the first move, the shortest and longest distances the man moves are;

d1 = √(85) and d2 = √(113)

Speed of the man, v = 0.5 cm/sec

Therefore by unit conversion function of a graphing calculator;

v = (25/1524) ft./sec

The times taken are;

t1 = √(85)/(25/1524) ≈ 562.023

t2 = √(113)/(25/1524) ≈ 648.014.

Average time, t = (t1 + t2)/2

Which gives;

t = (562.023 + 648.014)/2 ≈ 605.02

The average time it takes the man to escape in first move is t ≈ 605.02 seconds.

Learn more about Pythagorean theorem and probability here:

brainly.com/question/27180985

brainly.com/question/654982

brainly.com/question/24756209

#SPJ1

3 0

1 year ago