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3 years ago

Convert 3 5/8 + 1 3/4 to an improper fraction

Mathematics

2 answers:

Charra [1.4K]3 years ago

7 0

 ans is 43/8.

Hope this will help u...✳✴⤴⤴

fenix001 [56]3 years ago

7 0

3 $\frac{5}{8}$ + 1 $\frac{3}{4}$ = $\frac{3*8 + 5}{8}$ + $\frac{4+3}{4}$ = $\frac{29}{8}$ + $\frac{7}{4}$ = $\frac{29+14}{8}$ (We take 8, which is the LCM of 4 and 8, as the common denominator) = $\frac{43}{8}$

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Find x in this proportion. x/3 = x+2/5 x =

pickupchik [31]

$x = 3 \\ solution \\ \frac{x}{3} = \frac{x + 2}{5} \\ or \: 5x = 3(x + 2)(cross \: multiplication) \\ or \: 5x = 3x + 6 \\or \: 5x - 3x = 6 \\ or \: 2x = 6 \\ or \: x = \frac{6}{2} \\ x = 3 \\ hope \: it \: helps$

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

Find the volume for the regular pyramid. V =

mylen [45]

The volume is given by:
V = L * Apb * h
Where,
L: side of the base
Apb: apothema of the base
h: height
Substituting we have:
V = (4) * ((root (3) / 2) * 4) * (6)
Rewriting we have:
V = (4) * ((root (3) / 2) * 4) * (6)
V = 96 * (root (3) / 2)
V = 48 * root (3)
Answer:
V = 48 * root (3)

7 0

3 years ago

Read 2 more answers

Please Help!! and explain in detail!!!!

tatuchka [14]

Answer:

1/36 = 0.02777...

Step-by-step explanation:

Given the sequence 1, 0.25, 0.111..., 0.0625, 0.04, you would like to know the next term.

<h3>Terms</h3>

We recognize the decimal values to be equivalent to ...

1, 1/4, 1/9, 1/16, 1/25