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3 years ago

5 2/4 + 5 6/8 solve the fraction

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

11 2/8

Step-by-step explanation:

Turn to same fraction using multiplication

5 4/8 + 5 6/8

Add whole numbers and fractions sepperatly

5 + 5 + 4/8 + 6/8

10 + 10/8

10 + 1 2/8

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What two numbers multiplied equal -9 but add up to -6

jeka94

No numbers, is the question correct? Are you factoring?

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2 years ago

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Marina86 [1]

The first option, because if it’s 291 yards and 1/4 BELOW (below: meaning negative because you are under the usual sea level).

-291 (1/4)

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8 0

3 years ago

Write the number for nine million seven hundred thousand

marin [14]

Answer:

9,700,000.

Explain:

9 goes in the 1,000,000th place, and 700 would be placed right behind the 9's comma. Therefore, 9,700,000

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2 years ago

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Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

ZE is the angle bisector of AngleYEX and the perpendicular bisector of Line segment G F. Line segment G X is the angle bisector

frozen [14]

Answer:

Point A is the center of the circle that passes through points E, F, and G and the center of the circle that passes through points X, Y, and Z.

Step-by-step explanation:

A is the intersection of angle bisectors, so is the incenter of triangle EFG. It is also the intersection of the perpendicular bisectors of the sides of triangle EFG, so is the circumcenter.

The altitudes at X, Y, and Z are perpendicular to sides EF, EG, and FG, and pass through the incenter, so X, Y, Z are points on the incircle.

A is the center of circles through E, F, and G, and through X, Y, and Z.

6 0

3 years ago

Read 2 more answers