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3 years ago

Whats 80000000000000x200000004500

Mathematics

1 answer:

Sonbull [250]3 years ago

8 0

Answer:

80000000000000x200000004500 =

1.6e+25 which is essentially 1.6 with 25 0's after it

Step-by-step explanation:

May I have brainliest please? I'm trying to get 15 and I'm so close :)

You might be interested in

28 1/2 divided by 1/2

Alchen [17]

Answer: 57

Step-by-step explanation:

28 1/2 / 1/2

make 28 and 1/2 into 57/2

so now 57/2 divided by 1/2

the same as 57/2 times 2/1

multiply straight across

(57)(2)/(2)(1)

114/2

8 0

3 years ago

Read 2 more answers

I don't get # 2 please help and fast

Neporo4naja [7]

I wouldn't say that. It looks to me like you've absolutely got it ... almost.

A). Your fraction is correct. But it's not in simplest form like they want it.
Reduce 18/45 to simplest form. Hint: Divide top and bottom by 9 .

B). They want the percentage that's the same as 18/45 .
Do you remember how to change a fraction to percent ?
A fraction is just a short way to say "division".
" 18/45 " means " 18 divided by 45 ".
Do the division. To change the quotient into percent, multiply it by 100 .
(Same as moving the decimal point 2 places to the right.)
Round it the nearest whole number, if it isn't already a whole number.

C). Again, your fraction is correct, but it isn't in simplest form.
Reduce 27/45 to simplest form. Hint: Divide top and bottom by 9 again.

D). Change this fraction to percent, just like you did for the female fraction.
(Do the division that the fraction says, and multiply the quotient by 100.)
(Round to the nearest whole number percent if it isn't already.)

You already did the technical stuff. I just added some mechanical things.

5 0

3 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

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

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

Round this number to the given place values:

nataly862011 [7]

Answer:

Ones place: 456.3000

tenths place: 456.3000

hundredths place: 456.2990

thousandths place: 456.2991

5 0

3 years ago

-6 is greater than or equal to 10- 8x

Mamont248 [21]

Answer:

-6≥10-8x

8x≥16

x≥2

Step-by-step explanation:

8 0

4 years ago