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3 years ago

If x+3/3 = y+2/2, then x/3=

Mathematics

2 answers:

kotegsom [21]3 years ago

6 0

Answer:

C.) y/2

Step-by-step explanation:

We can simplify (x+3)/3 to x/3 + 3/3, then simplify (y+2)/2 to y + 2/2, and since 2/2 and 3/3 are both 1, so we can say that x/3 + 1 = y/2 + 1, meaning that x/3 = y/2.

olga55 [171]3 years ago

4 0

Answer:

c. y/2

Step-by-step explanation:

We can solve this by cross-multiplying.

1. Since $\frac{x+3}{3}$ = $\frac{y+2}{2}$ , we can cross-multiply both sides of the equation to remove the denominators. Multiple the right side by 2 and the left side by 3 to get 2(x+3)=3(y+2).

2. Distribute to get the x on one side. 2x+6=3y+6.

3. The 6s on both sides cancel out so we are left with 2x=3y.

4. Divide both sides by 2 to get x on one side, so x= $\frac{3y}{2}$ .

5. The question asks for x/3, so we can divide both sides of the equation to get $\frac{x}{3}$ = $\frac{3y}{6}$ .

6. Simplify. We get $\frac{x}{3}$ = $\frac{y}{2}$ , so c is the answer.

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3 years ago

Read 2 more answers

LINEAR ALGEBRA

kenny6666 [7]

Answer:

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

Step-by-step explanation:

Let be $\vec u_{1} = [2,3,1]$ , $\vec u_{2} = [4,1,0]$ and $\vec u_{3} = [1, 2,k]$ , $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{3}$ if and only if:

$\alpha_{1} \cdot \vec u_{1} + \alpha_{2} \cdot \vec u_{2} +\alpha_{3}\cdot \vec u_{3} = \vec O$ (Eq. 1)

Where:

$\alpha_{1}$ , $\alpha_{2}$ , $\alpha_{3}$ - Scalar coefficients of linear combination, dimensionless.

By dividing each term by $\alpha_{3}$ :

$\lambda_{1}\cdot \vec u_{1} + \lambda_{2}\cdot \vec u_{3} = -\vec u_{3}$

$\vec u_{3}=-\lambda_{1}\cdot \vec u_{1}-\lambda_{2}\cdot \vec u_{2}$ (Eq. 2)

$\vec O$ - Zero vector, dimensionless.

And all vectors are linearly independent, meaning that at least one coefficient must be different from zero. Now we expand (Eq. 2) by direct substitution and simplify the resulting expression:

$[1,2,k] = -\lambda_{1}\cdot [2,3,1]-\lambda_{2}\cdot [4,1,0]$