Solve variable 3/4x-5=7

Write a 3-digit number that can be divided by 3 but not by 10.

NARA [144]

Answer:

Hi

999

Step-by-step explanation:

8 0

3 years ago

For their twentieth wedding anniversary, Richard and Sylvia had a party for several of their best friends. Things were going alo

kompoz [17]

Answer:

Richard is older

Step-by-step explanation:

We can set up an inequality for both statements.

Let r equal Richard's age and s equal Sylvia's age.

"I am older than my wife."

Since Richard is speaking, the inequality would look like this:

r > s

This means Richard is older than Sylvia.

"I am younger than my husband."

Since Sylvia is speaking, the inequality would look like this:

s < r

This means that Sylvia is younger than Richard.

We can flip one inequality to "see" them from the same perspective.

Let's use s < r

To make it so that we can see the relationship from Richard's perspective, flip the entire inequality.

s < r

to

r > s

The inequality from the first quote is identical to this one!

Therefore, Richard is older than Sylvia.

8 0

3 years ago

Find the unlabeled side length

White raven [17]

Answer:

hope it helps you.......

8 0

3 years ago

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Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago

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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