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3 years ago

1,432 ×54 multipy to find the product.Thanks

Mathematics

2 answers:

TiliK225 [7]3 years ago

4 0

The answer is 1432 x 54 = 77,328

faust18 [17]3 years ago

3 0

1,432 x 54 = 77,328

Have a good day! -Wajiha

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Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. C = 71°, a = 27, c = 26

guajiro [1.7K]

We have

SinC/ c = Sin A / a

Sin 71/ 26 = Sin A / 27

Sin A = 27 Sin 71 / 26 = about .982

So°

Sin-1(.982) = A = 79. 08°

Then angle B = 180 - 71 - 79.08 = 29.92°

And b is given by

b/sin29.92 = 26/sin 71

b = 26sin29.92/sin71 = about 13.72

But A could also be an obtuse angle = 180 - 79.08 = 100.92°

So we have

B = 180 - 71 - 100.92 = 8.08°

And we have

b / sin 8.08 = 26/sin71

b = 26sin8.08/sin 71 = 3.865

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3 years ago

Jorge's hourly salary is $7.65. last week he worked 23 hour week how much did he earn

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$7.65 × 23 = $175.95

Answer: $175.95

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2 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.

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2 years ago

Pls help this is an ixl question and it really is hard

valentina_108 [34]

Answer:

It would be the first one... 10:25

Step-by-step explanation:

10/4=2.5

2.5*10=25

2.5*4=10

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2 years ago

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45 min. - 4 peoples - 5 walls
? min. - 6 peoples - 4 walls
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36 min. - 1 man - 1 wall ⇒
6 min. - 6 peoples - 1 wall ⇒
24 min. - 6 peoples - 4 walls (answer)

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3 years ago