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steposvetlana [31]
3 years ago
13

Which of the following is an example of a system of three linear equations in theee variables

Mathematics
2 answers:
harkovskaia [24]3 years ago
6 0

Answer:

I think option D is correct

Gnom [1K]3 years ago
5 0

Answer:

1) no, because a solution must be an ordered triple pair (x,y,z)

2) { x+y+z=5    

   { 2x-3y+z=7      

   { x+2y-4z=2

3) infinity many solutions

4) use equations (1) and (2) to eliminate y

5) solve one equation for one of its variables  

Step-by-step explanation:

should be 100% on the quick check if you put those answers

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Need help on questions 1, and 2 please
finlep [7]
1. See the attached graph.

2. All translations are considered rigid motion. The locations of points on the figure do not change relative to one another.

7 0
3 years ago
Explain why the expression 10 to the power of zero, one to the power of four and 1×1.0 to the power of zero have the same value
Nina [5.8K]

Answer:

10*0=1, according to the law of indices which states that anything to the zero power is equal to one

1*4=1×1×1×1=1

1×1.0=1.0=1

3 0
3 years ago
Fatima has made a planter box that is 2.5 feet long, 1.5 feet wide, and 2 feet high.
earnstyle [38]

Answer:

Bottom right

First you multiply the bottom (2.5 x 1.5) to get 3.75. Then multiply 3.75 by 2 to got 7.5.

Whole equation: (2.5 x 1.5) x 2.0 = 7.5

3 0
3 years ago
Roots of Quadratics 50 PTS!!!
noname [10]

The values of k for the different quadratic equation solutions are as follows

a  the equation 2x² - x + 3k = 0 has two distinct real roots

  • k < 1/24

b. the equation 5x² - 2x + (2k − 1) = 0 has equal roots

  • k = 3/5

ci the equation -x² + 3x + (k + 1) = 0 has real roots

  • k > -3.25

d the equation 3kx² - 3x + 2 = 0 has no real solutions

  • k < ± 1.633

<h3>How to solve quadratic equations to get different answers</h3>

Quadratic equations of the form ax² + bx + c = 0 is solved using the formula

-b+\frac{\sqrt{b^{2}-4ac } }{2a}     OR     -b-\frac{\sqrt{b^{2}-4ac } }{2a}

The equation b² - 4ac is called the discriminant and it is used as follows

To solve the equation and get two real roots: 2x² - x + 3k = 0

  • b² - 4ac > 0

substituting the values gives

(-1)² - 4 * 2 * 3k > 0

1 - 24k > 0

1 > 24k

divide through by coefficient of k

k < 1/24

To solve the equation and get equal roots: 5x² - 2x + (2k − 1) = 0

  • b² - 4ac = 0

substituting the values gives

(-2)² - 4 * 5 * (2k - 1) = 0

4 - 40k + 20 = 0

-40k = -24

divide through by coefficient of k

k = 3/5

To solve the equation and get real roots  -x² + 3x + (k + 1) = 0

  • b² - 4ac > 0

substituting the values gives  

(3)² - 4 * -1 * (k+1) > 0

9 + 4k + 4> 0

4k > -13

divide through by coefficient of k

k > -3.25

To solve the equation and get  no real solutions  3kx² - 3x + 2 = 0

  • b² - 4ac < 0

substituting the values gives  

(-3)² - 4 * 3k * 2 < 0

9 - 24k² > 0

9 > 24k²

divide through by coefficient of k²

k² < 24/9

k < ± 1.633

Learn more about roots of quadratic equations: brainly.com/question/26926523

#SPJ1

4 0
1 year ago
Read 2 more answers
How do you solve this equation . Q=3a+5ac
mr_godi [17]
A "solution" would be a set of three numbers ... for Q, a, and c ... that
would make the equation a true statement.

If you only have one equation, then there are an infinite number of triplets
that could do it.  For example, with the single equation in this question,
(Q, a, c) could be (13, 1, 2) and they could also be  (16, 2, 1).
There are infinite possibilities with one equation.

In order to have a unique solution ... three definite numbers for Q, a, and c ...
you would need three equations.
8 0
3 years ago
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