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Black_prince [1.1K]

3 years ago

15

At a family reunion, a family friend brings 20 pounds of ice and 30 cups. Which expression shows the greatest common factor, usi

ng the Distributive Property, of 20 and 30 ?

1 answer:

Eddi Din [679]3 years ago

6 0

The expression for the greatest common factor of 20 and 30 using the distributive property is 10(2 + 3)

Pounds of ice = 20

Number of cups = 30

The relation for the distributive property :

a × (b + c)

Expand ;

a × (b + c) = (a × b) + (a × c) = ab + ac

Finding the greatest common factor of 20 and 30 ;

20 = 2 × 10

30 = 3 × 10

(2 × 10) + (3 × 10)

According to the distributive property

(2 × 10) + (3 × 10) = 10(2 + 3)

Therefore, the expression for the greatest common factor is 10(2 + 3)

Learn more :brainly.com/question/15263211

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Step-by-step explanation:

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Which expression is equivalent to the following complex fraction? X+5/x+2-x+1/x^2+2x

ICE Princess25 [194]

Answer:

Final answer will be the choice which matches best with expression

$\frac{x^2+4x-1}{x\left(x+2\right)}$ or

$\frac{x^2+4x-1}{x^2+2x}$

Step-by-step explanation:

Given expression is:

$\frac{\left(x+5\right)}{\left(x+2\right)}-\frac{\left(x+1\right)}{\left(x^2+2x\right)}$

We begin by factoring denominators:

$=\frac{\left(x+5\right)}{\left(x+2\right)}-\frac{\left(x+1\right)}{x\left(x+2\right)}$

Multiply and divide first term by (x) to make denominators equal.

$=\frac{x\left(x+5\right)}{x\left(x+2\right)}-\frac{\left(x+1\right)}{x\left(x+2\right)}$

Since denominators are equal so we can combine numerators.

$=\frac{x\left(x+5\right)-\left(x+1\right)}{x\left(x+2\right)}$

Now simplify

$=\frac{x^2+5x-x-1}{x\left(x+2\right)}$

$=\frac{x^2+4x-1}{x\left(x+2\right)}$

$=\frac{x^2+4x-1}{x^2+2x}$

Hence final answer will be the choice which matches best with expression

$\frac{x^2+4x-1}{x\left(x+2\right)}$ or

$\frac{x^2+4x-1}{x^2+2x}$

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

This is solving differential equations by sepeartion of variables

Brums [2.3K]

$\mathrm dy+k(y-70)\,\mathrm dx=0$

Separating variables gives

$\dfrac{\mathrm dy}{y-70}=-k\,\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dy}{y-70}=-k\int\mathrm dx$
$\ln|y-70|+C_1=-kx+C_2$
$\ln|y-70|=-kx+C$

Solving for $y$ gives

$e^{\ln|y-70|}=e^{-kx+C}$
$y-70=e^{-kx}e^C$
$y=Ce^{-kx}+70$

As $y(1)=140$ , you get

$140=Ce^{-k}+70$
$70=Ce^{-k}$
$C=70e^k$

so the particular solution to this ODE is

$y=70e^ke^{-kx}+70$
$y=70e^{k(1-x)}+70$

Unless you have any more information, you're done.

5 0

3 years ago