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kolezko [41]
3 years ago
15

Find a system of two equations in two variables, x1 and x2, that has the solution set given by the parametric representation x1

= t and x2 = 5t − 6, where t is any real number. (Enter your answer as a comma-separated list of equations.)
Mathematics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

The required system of equations to the given parametric equations are:

5x1 - x2 = 6

x1 + x2 = -6

Step-by-step explanation:

Given the parametric equations:

x1 = t

x2 = -6 + 5t

Eliminating the parameter t, we obtain one of the equations of a system in two variables, x1 and x2 that has the solution set given by the parametric equations.

Doing that, we have:

5x1 - x2 = 6

Again a second equation can be a linear combination of x1 and x2

x1 + x2 = -6 + 6t

x1 + x2 = -6 (putting t=0)

And they are the required equations.

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r-ruslan [8.4K]

Answer:

Gene has 45 stamps

Step-by-step explanation:

7 0
2 years ago
<img src="https://tex.z-dn.net/?f=3y%20-%205x%20%3D%2010" id="TexFormula1" title="3y - 5x = 10" alt="3y - 5x = 10" align="absmid
kompoz [17]

Answer:

\dfrac{5}{3}

Step-by-step explanation:

To find the gradient (slope) of the given <u>linear relation</u>, use <u>arithmetic operations</u> to isolate y:

Given relation:

3y-5x=10

Add 5x to both sides:

\implies 3y-5x+5x=10+5x

\implies 3y=5x+10

Divide both sides by 3:

\implies \dfrac{3y}{3}=\dfrac{5}{3}x+\dfrac{10}{3}

\implies y=\dfrac{5}{3}x+\dfrac{10}{3}

<u>Slope-intercept form of a linear equation</u>

y = mx+b

where:

  • m is the slope (gradient)
  • b is the y-intercept

Comparing the rewritten equation with the slope-intercept formula, the gradient (slope) of the given linear relation is ⁵/₃.

Learn more about linear equations here:

brainly.com/question/27317293

brainly.com/question/27781455

7 0
1 year ago
Read 2 more answers
Solve for x: 6x-4 over 3+2x=5/6
viktelen [127]

Domain:\\\\3+2x\neq0\to x\neq-1.5\\\\\dfrac{6x-4}{3+2x}=\dfrac{5}{6}\qquad\text{cross multiply}\\\\6(6x-4)=5(3+2x)\qquad\text{use distributive property}\\\\(6)(6x)+(6)(-4)=(5)(3)+(5)(2x)\\\\36x-24=15+10x\qquad\text{add 24 to both sides}\\\\36x=39+10x\qquad\text{subtract 10x from both sides}\\\\26x=39\qquad\text{divide both sides by 26}\\\\x=\dfrac{39}{26}\\\\x=\dfrac{39:13}{26:13}\\\\\boxed{x=\dfrac{3}{2}}\in D

5 0
3 years ago
Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?
alexgriva [62]

Answer:

x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)

OR x^9/(∛y)

Step-by-step explanation:

Given the indicinal equation

(x^27/y)^1/3

To find the corresponding expression, we will simplify the equation as shown:

(x^27/y)^⅓

= (x^27)^⅓/y⅓

= {x^(3×9)}^⅓/y⅓

= x^9/y⅓

= x^9/(∛y)

The right answer is x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)

6 0
3 years ago
Read 2 more answers
Twelve is added to the product of a number and 5. The result is -3. Find the number
TiliK225 [7]
5 + -8 = -3.

So the answer would be -8.
4 0
3 years ago
Read 2 more answers
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