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3 years ago

Will someone help me with both of these I have to solve for the unknown portions please

Mathematics

2 answers:

Kobotan [32]3 years ago

6 0

Answer:

h = 5/2
p = 6

Step-by-step explanation:

c) Multiply by the denominator under the variable, then simplify.

$\dfrac{10}{8}=\dfrac{h}{2}\\\\\dfrac{2\cdot 10}{8}=h\\\\\dfrac{5}{2}=h$

d) Simple observation shows you the denominator of each fraction is half the numerator, so p = 6.

If you want to solve this algebraically, you can multiply by p and divide by the value on the left of the equal sign (2).

$\dfrac{4}{2}=\dfrac{12}{p}\\\\\dfrac{p}{2}\cdot\dfrac{4}{2}=\dfrac{p}{2}\cdot\dfrac{12}{p}\\\\p=6$

For an "upside down" proportion such as this (with the variable in the denominator), you can rewrite it by inverting both fractions:

$\dfrac{2}{4}=\dfrac{p}{12}$

Now, solve as for (c), by multiplying by the denominator under the variable:

$\dfrac{12\cdot 2}{4}=p=6$

_____

<em>Comment on this last solution method</em>

In general, you can use equation solving methods to solve inequalities. However, taking a reciprocal is an order-reversing process:

3 > 2

1/3 < 1/2

So, if you want to solve an inequality with the variable in the denominator of a proportion, be careful to observe any order-changing behavior of your solution method.

IceJOKER [234]3 years ago

3 0

Answer:

b=2.5

p=6

Step-by-step explanation:

Take the fraction and simplify it. So 10/8 would be 5/4

You multiply the 2 by 2, giving you h/4

So now you have 5/4=h/4

h=5 but now you have to divide both by 2 again so it can go back to the original fraction.

5÷2=2.5 and 4÷2=2

Second Problem! 4/2 is equal to 2/1

You can also multiply 4 by 3 to get the numerator as 12. Multiply the 4 by 3 and the 2 by 3

Now you have 12/6= 12/p

So p is 6

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2 years ago

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3 years ago

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3 years ago

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