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3 years ago

Explain in numbers: 22% of 536

1 answer:

ivann1987 [24]3 years ago

6 0

So in order for you to get 22% of first you must subtract .22 from 1.00 to get .78 then you multiply 573 by .78 to get your answer. Or you can just divided 573 by 100 then multiply by 22 and subtract the outcome of the from 573

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HURRY PLZ

Oliga [24]

Answer:

6x + 2(−2x + 1) = 22

Step-by-step explanation:

6x + 2(−2x + 1) = 22

This step can be used to solve by substitution method.

8 0

3 years ago

Pic is question vuygvcutyfvgt

Furkat [3]

Answer:

Step-by-step explanation:

270/9 = 30

Hope This Helps & Good Luck :)

5 0

3 years ago

Write a decimal that is 1/10 of .9

amm1812

1/10 of .9 is the same as dividing .9 by 10. .90 / 10 = .09 .09 is the answer

8 0

3 years ago

Which value of a makes the equation 13=40-3a

weeeeeb [17]

The value of A is 9. This is because 40-13= 27
3x?=27 3x9=27

3 0

4 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago