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Sever21 [200]
2 years ago
12

State all integer values of x in the interval (2,8) that satisfy the following inequality:

Mathematics
1 answer:
lidiya [134]2 years ago
3 0

Answer:

3,4,5,6 and 7

Step-by-step explanation:

Given

5x + 4 \ge 19

Required

State the values of x within 2 and 8

5x + 4 \ge 19

Collect Like Terms

5x \ge 19 - 4

5x \ge 15

Divide both sides by 5

\frac{5x}{5} \ge \frac{15}{5}

x \ge 3

(2,8) is an open interval which means that 2 and 8 are not inclusive of the values of x.

So:

The values of x \ge 3 are: 3,4,5,6 and 7

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3 years ago
Need helpppppppppp please
aleksley [76]
Adding the two equations
3x + 6y + 3x - 6y = 36+0
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subtracting second equation from first
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8 0
2 years ago
Read 2 more answers
6) (after 2.3) Consider the infinite system of linear equations in two variables given by ax + by = 0 where (a, b) moves along t
Paraphin [41]

Answer:

Step-by-step explanation:

Given infinite system of linear equations is ax + by = 0

when (a,b) moves along unit circle in plane.

a) system having unique system (0, 0)

Since two of equation in thus system will be

1.x+0.y=0\\x=0

and

0.x+1.y=0\\y=0

It is clear that x = 0, y= 0 is the only solution

b) Linear independent solution in this system gives some set of solutions

1.x+0.y=0\\\x=0

and

0.x+1.y=0\\y=0

Vector form is

\left[\begin{array}{ccc}1&0\\0&1\end{array}\right] =I

c) for this equation if add 0x +0y = 0 to system , Nothing will change

Because [0,0] satisfies that equation

d) If one of the equation is ax + by = 0.00001

where 0.00001 is small positive number

so, the system will be inconsistent

Therefore, the system will have no solution.

6 0
3 years ago
In AABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of
sergij07 [2.7K]

Answer:

<h3>35</h3><h3 />

Step-by-step explanation:

see attached image.

perimeter of trapezoid = sides (6 + 8 + 5 + 8 + 8)

                                     = 35

8 0
3 years ago
Can someone help me :
morpeh [17]
E is the answer I am pretty sure.
4 0
2 years ago
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