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3 years ago

Solve my problem, PLEASE!!!!

Mathematics

2 answers:

N76 [4]3 years ago

3 0

Y=-1/3x+5
explanation

notka56 [123]3 years ago

3 0

Answer:

y=-3+15

Step-by-step explanation:

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A scientist needs 4.8 liters of a 12% alcohol solution. She has available a 22% and a 10% solution. How many liters of the 22% a

yanalaym [24]

Answer:

4.0 L of 10%
0.8 L of 22%

Step-by-step explanation:

For mixture problems, it is convenient to define a variable to represent the amount of the greatest contributor. Let x represent the amount of 22% solution in the mix. Then 4.8-x is the amount of 10% solution.

The amount of alcohol in the mix is ...

0.22x +0.10(4.8-x) = 0.12(4.8)

Eliminating parentheses, we have ...

0.22x -0.10x +0.10(4.8) = 0.12(4.8)

Subtracting (0.10)(4.8) and combining x-terms gives ...

0.12x = 0.02(4.8)

x = (0.02/0.12)(4.8) = 0.8 . . . . . divide by the x-coefficient

The scientist needs 0.8 L of 22% solution and 4.0 L of 10% solution.

5 0

3 years ago

Three consecutive integers can be represented by n, n + 1, and n + 2. if the seum of three consecutive integers is 57, what are

Helen [10]

Sum means add
sum of the 3 is
n+n+1+n+2

it add to 57
n+n+1+n+2=57
3n+3=57
minus 3 both sides
3n=54
divide both sides by 3
n=15
n+1=16
n+2=17

the integers are 15,16, and 17

8 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

What is 500 x 50, 000

kogti [31]

Answer is 25,000,000

6 0

3 years ago

What is the integral of a square root? Please explain.

IrinaK [193]

If y = xⁿ

∫y dx = xⁿ⁺¹ / (n + 1) + C Provided n ≠ -1.

y = √x

y = x^(0.5)

∫y dx = x^(0.5+1) / (0.5 + 1) = x^(1.5) / 1.5 = x^(1.5) / (3/2)

∫y dx = (2/3) x^(1.5) + C.

∫y dx = (2/3) x^(3/2) + C.

∫y dx = (2/3)√x³ + C

5 0

3 years ago

Read 2 more answers