Okay, After this i'm done asking for help. I hope..

translate the sentence into a equation using n as the unknown number . Then solve the equation for n . 5 increased by half a num

Ksju [112]

Answer:

I think it'll be 5 + n = 11

11 - 5 = 6

n = 6

5 0

3 years ago

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Maricella solves for x in the equation 4x-2(3x-4)+4=-x+3(x+1)+1. She begins by adding –4 + 4 on the left side of the equation an

Elodia [21]

Maricella’s strategy is incorrect as the values added on the left and right side of the equation changes the original equation $4x-2(3x-4)+4=-x+3(x+1)+1$ to $4x-2(3x-4)+4=-x+3(x+1)+3$ .

Further explanation:

Maricella begin by adding $-4+4$ on the left side of the equation and $1+1$ on the right side of the equation $4x-2(3x-4)+4=-x+3(x+1)+1$ .

The value added on the left is equal to zero so it does not change the equation but the value added on the right is 2, a non-zero digit so it changes the equation as shown below.

$4x-2(3x-4)+4-4+4=-x+3(x+1)+1+1+1\\4x-2(3x-4)+4+0=-x+3(x+1)+1+2\\4x-2(3x-4)+4=-x+3(x+1)+3$

Therefore, the strategy used by Maricella will change the equation and so the solution of the equation will be incorrect.

One way of solving such a linear equation is by adding or subtracting the required value on each side of the equation. and then simplifying the further equation.

Similarly, the other way to solve any linear equation is by separating the variable terms on one side of the equation and constant terms on the other side of the equation.

After this simplify the linear equation to obtain the solution of the equation for $x$ .

Whereas Maricella’s strategy fails to solve the equation $4x-2(3x-4)+4=-x+3(x+1)+1$ since the values added and subtracted in the equation changes the original equation completely.

Learn more:

1. Linear equation application brainly.com/question/2479097

2. To solve one variable linear equation brainly.com/question/1682776

3. Binomial and trinomial expression brainly.com/question/1394854

Answer details

Grade: Middle school

Subject: Mathematics

Chapter: Linear equation

Keywords: equation, linear equation, Maricella, right side, left side, strategy, solve for $x$ , adding, subtracting, variable, constant, solution of the equation, value, original equation, variable terms, constant terms.

4 0

2 years ago

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I need help on finding an equation!?

Sedbober [7]

Your distance from Seattle after two hours of driving at 62 mph, from a starting point 38 miles east of Seattle, will be (38 + [62 mph][2 hr] ) miles, or 162 miles (east).

Your friend will be (20 + [65 mph][2 hrs] ) miles, or 150 miles south of Seattle.

Comparing 162 miles and 150 miles, we see that you will be further from Seattle than your friend after 2 hours.

After how many hours will you and your friend be the same distance from Seattle? Equate 20 + [65 mph]t to 38 + [62 mph]t and solve the resulting equation for time, t:

20 + [65 mph]t = 38 + [62 mph]t

Subtract [62 mph]t from both sides of this equation, obtaining:

20 + [3 mph]t = 38. Then [3 mph]t = 18, and t = 6 hours.

You and your friend will be the same distance from Seattle (but in different directions) after 6 hours.

7 0

3 years ago

What fraction is equivalent to negative 3 over 2

Llana [10]

Answer:

-6/4

Step-by-step explanation:

5 0

3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

<u>Given equation</u>:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

4 0

2 years ago