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2 years ago

Please helpppppppppp

Mathematics

1 answer:

amm18122 years ago

7 0

Your answer is A) Acute

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Kaylee drove 75.5 miles in 54 minutes,if she continues to drive at the same rate, how long will it take her to drive 250 miles?

Crazy boy [7]

Answer:

179 minutes

Step-by-step explanation:

Divide both sides by 75.5 to see how many minutes it would take to drive 1 mile. (54 / 75.5) =108/151
She can drive 1 mile in 108/151 minutes
Times by 250. (108/151 * 250) 178.8
179 to nearest minute

7 0

3 years ago

What is the form of the Pythagorean triple generator?

stepladder [879]

Answer:(x^2+y^2)^2=(x^2+y^2)(x^2+y^2)

Step-by-step explanation:

We can rewrite left side into right side form

(x^2+y^2)^2=(x^2+y^2)(x^2+y^2)

we can expand it

(x^2+y^2)^2=x^4+x^2y^2+x^2y^2+y^4

(x^2+y^2)^2=x^4+y^4+2x^2y^2

we can add and subtract 2x^2y^2

(x^2+y^2)^2=x^4+y^4+2x^2y^2+2x^2y^2-2x^2y^2

(x^2+y^2)^2=x^4-2x^2y^2+y^4+2x^2y^2+2x^2y^2

(x^2+y^2)^2=x^4-2x^2y^2+y^4+4x^2y^2

(x^2+y^2)^2=x^4-2x^2y^2+y^4+(2xy)^2

(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2

4 0

2 years ago

Read 2 more answers

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

2 years ago

100 point to the first correct answer

miss Akunina [59]

Answer:

X ~ B(n, p)

Given: X ~ B(2, 5/17)

⇒ n = 2

⇒ p = 5/17

Binomial distribution formula:

$\sf P(X = x) = nC_x (1-p)^{(n-x)}p^x$