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1 year ago

Apply the distributive property to factor out the greatest common factor. 12 + 20 =

Mathematics

1 answer:

guajiro [1.7K]1 year ago

6 0

We can use prime factorization to find the GCF:

$12 = 3^1 \cdot 4^1$

$20 = 4^1 \cdot 5^1$

The greatest value both 12 and 20 share is 4. By factoring this out of the equation, we get:

4(3 + 5) = 4(8) = 32

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silvia compro la misma licuadora que daniela, que costaba $355.5, pero como abonó con tarjeta de credito, le recargaron un 16%,

Debora [2.8K]

Answer:

Costo final= $412.38

Step-by-step explanation:

Dada la siguiente información:

Costo inicial= $355.5

Recargo de la tarjeta= 16% = 0.16

<u>Para calcular el costo final que debe pagar Silvia, debemos usar la siguiente información:</u>

Costo final= costo inicial*(1 + recargo)

Costo final= 355.5*1.16

Costo final= $412.38

4 0

1 year ago

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

3 0

1 year ago

Squaring both sides of an equation is irreversible. Is Cubing both sides of an equation reversible? Provide numerical examples t

nikitadnepr [17]

Answer:

<h2>Cubing both sides of an equation is reversible.</h2>

Step-by-step explanation:

Squaring both sides of an equation is irreversible, because the square power of negative number gives a positive result, but you can't have a negative base with a positive number, given that the square root of a negative number doesn't exist for real numbers.

In case of cubic powers, this action is reversible, because the cubic root of a negative number is also a negative number. For example

$\sqrt[3]{x} =-1$