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2 years ago

Please hurry simplify the expression (15^1/5)^5

Mathematics

1 answer:

Lelu [443]2 years ago

5 0

Answer:

Step-by-step explanation:

using the rule of exponents

$(a^m)^{n}$ = $a^{mn}$

$(15^{\frac{1}{5)} } ^{5}$ = $15^{(1/5(5))}$ = $15^{1}$ = 15

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one half multiplyed by 4 pluse one quarter times 4 pluse three quarters times 4 pluse three 8s times. 3 pluse 5 8s times 3

elena-14-01-66 [18.8K]

.5 times four is 3 plus one quarter is 2.33 times 4 is 9.32 plus 75 is 10.07plus 24 is 24.07 then 32 and answer is 56.07

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3 years ago

36% of a number is 63.Find 124% of the number

Trava [24]

Answer:

217

Step-by-step explanation:

124%=x

36%=63

=> $x=\dfrac{63\times 124}{36}=217$

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3 years ago

If there is a triangle and the sides are 8&12 and 14 what is the missing angle

rewona [7]

Answer:

8/12 = 14/x

Cross multiplication so

8x=14×12

8x=168

X=168/8

X=21

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2 years ago

Determine the values of the constants r and s such that i(x, y) = x rys is an integrating factor for the given differential equa

garri49 [273]

$\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0$

For the ODE to be exact, we require that $M_y=N_x$ , which we'll verify is not the case here.

$M_y=21xy^2+6$
$N_x=2xy^2-1$

So we distribute an integrating factor $i(x,y)$ across both sides of the ODE to get

$iM\,\mathrm dx+iN\,\mathrm dy=0$

Now for the ODE to be exact, we require $(iM)_y=(iN)_x$ , which in turn means

$i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM$

Suppose $i(x,y)=x^ry^s$ . Then substituting everything into the PDE above, we have

$x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)$
$19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s$
$19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s$
$\implies\begin{cases}r-7s=19\\r+6s=-7\end{cases}\implies r=5,s=-2$

so that our integrating factor is $i(x,y)=x^5y^{-2}$ . Our ODE is now

$(7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0$

Renaming $M(x,y)$ and $N(x,y)$ to our current coefficients, we end up with partial derivatives

$M_y=7x^6-6x^5y^{-2}$
$N_x=7x^6-6x^5y^{-2}$

as desired, so our new ODE is indeed exact.

Next, we're looking for a solution of the form $\Psi(x,y)=C$ . By the chain rule, we have

$\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)$

Differentiating with respect to $y$ yields

$\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}$
$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

Thus the solution to the ODE is

$\Psi(x,y)=x^7y+x^6y^{-1}=C$

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3 years ago

Use the expression to find the value of the 20th term.<br> A. 24<br> B. 76<br> C. 80<br> D. 84

Oliga [24]

Answer:

Step-by-step explanation:

if its talking about what ones a factor or moltiple of 20 then c if not then i dont know

5 0

3 years ago