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3 years ago

¿Se puede distribuir el exponente de una suma o una resta? ¿Y la raiz?

Mathematics

1 answer:

azamat3 years ago

6 0

Si se puede distribuir el exponente de una suma porque si 2+3 elevado al 2 entonces el Raiz seria cuadrada de 19

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Please answerrrr Graph g(x), where f(x) = 4x − 2 and g(x) = f(x + 1). a line labeled g of x that passes through points 0, negati

Mrrafil [7]

Answer:

a line labeled g of x that passes through points negative 1, negative 2 and 0, 2

Step-by-step explanation:

we have

$f(x)=4x-2$

$g(x)=f(x+1)$

To determine g(x), substitute the variable x by (x+1) in the function f(x)

$f(x+1)=4(x+1)-2$

$f(x+1)=4x+4-2$

$f(x+1)=4x+2$

$g(x)=f(x+1)$

$g(x)=4x+2$

therefore

a line labeled g of x that passes through points negative 1, negative 2 and 0, 2

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3 years ago

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Jason received $88 for his birthday. He went to a sporting good store and bought a baseball glove, baseball, and bat. He had $25

rusak2 [61]

Answer:

Step-by-step explanation:

88 - 25 = 63 did it on paper

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3 years ago

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Find the range of this function

jeyben [28]

Answer:

Not sure!

Step-by-step explanation:

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3 years ago

israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa

agasfer [191]

Lets solve our radical equation $\sqrt{x+6}-4=x$ step by step.

Step 1 add 4 to both sides of the equation:

$\sqrt{x+6} -4+4=x +4$

$\sqrt{x} +6=x+4$

Step 2 square both sides of the equation:

$\sqrt{x+6} =x+4$

$\sqrt{x+6}^2 =(x+4)^2$

$x+6=(x+4)^2$

Step 3 expand the binomial in the right hand side:

$x+6=x^2+8x+16$

Step 4 simplify the expression:

$0=x^2+8x-x+16-6$

$x^2+7x+10=0$

Step 5 factor the expression:

$(x+2)(x+5)=0$

Step 6 solve for each factor:

$x+2=0$ or $x+5=0$

$x=-2$ or $x=-5$

Now we are going to check both solutions in the original equation to prove if they are valid:

For $x=-2$

$\sqrt{x+6}-4=x$

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$

The solution $x=2$ is a valid solution of the rational equation $\sqrt{x+6}-4=x$ .

For $x=-5$

$\sqrt{x+6}-4=x$

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3\neq -5$

Since -3 is not equal to -5, the solution $x=-5$ is not a valid solution of the rational equation $\sqrt{x+6}-4=x$ ; therefore, $x=-5$ is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0

3 years ago

Sophie pays m dollars to rent a car for the weekend. On the drive home, she buys $25 worth of gasoline. Her friend melody pay fo

zalisa [80]

Answer:

25-1/2=

Step-by-step explanation:

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3 years ago