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2 years ago

A point (x, y) is the solution for the inequality y ≥ 2x + 1.

Mathematics

1 answer:

Radda [10]2 years ago

7 0

Answer:

(0,1)

Step-by-step explanation:

We are given that a point (x,y) is the solution for the inequality

$y\geq 2x+1$

We have to find ordered pair for which y has a value that is less than or equal to and satisfies the inequality shown.

a.(3,8)

Substitute x=3 and y=8

$8\geq 2(3)+1=7$

y has a value that is not less than or equal to 1.

b.(1,1)

Substitute x=1

$y\geq 2(1)+1=3$

But given y=1 which is not greater than or equal to 3.

Therefore,y has a value that is not less than or equal to 1.

c.(2,3)

Substitute x=2

$y\geq 2(2)+1=5$

3 is not greater than or equal to 5.

It is false.

d.(0,1)

Substitute x=0

$y2(0)+1=1$

y has a value which is equal to 1.

Hence, (0,1) is true option.

Answer:(0,1)

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Answer:

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2 years ago

Is the work shown below correct? Explain your answer.

riadik2000 [5.3K]

(11 + 21) - ( 3 - 100 )
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3 years ago

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

Let

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ = A $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = X and $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$ = B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A= $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$ $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}42\\0\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\0\\\end{array}\right]$

From here x= -3 and y= 0

Solution Set = [(-3,0)]

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2 years ago