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3 years ago

Write a linear inequality in two variables that has (-1, 3) as a solution but does not have (4,0) as a solution.

Mathematics

1 answer:

Stels [109]3 years ago

3 0

Answer: y ≤ -x + 2

Step-by-step explanation:

There are an infinite number of possible solutions.

Plot the two points, draw any line between them such that the two points are NOT connected, then choose the inequality symbol that shades (-1, 3).

Here is one solution that satisfies the requirements. y ≤ -x + 2

Here is another: y < -x + 3

And another: y > x + 3

And one more: y ≥ x + 4

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Sustituir. 20c + 15r + 75p + 50e

AVprozaik [17]

Answer:

8c - 7r + 10p - 20e

Step-by-step explanation:

Subtract (20c + 15r + 75p + 50e) - (12c + 22r + 65p + 70e)

(20c + 15r + 75p + 50e) - (12c + 22r + 65p + 70e)

Open bracket

20c + 15r + 75p + 50e - 12c - 22r - 65p - 70e

Collect like terms

= 20c - 12c + 15r - 22r + 75p - 65p + 50e - 70e

= 8c - 7r + 10p - 20e

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3 years ago

What is the remainder when (3x4 + 2x3 − x2 + 2x − 19) ÷ (x + 2)? 0 5 10 15

Elanso [62]

Answer: Second option is correct.

Explanation:

Since we have given that

$f(x)=3x^4+2x^3-x^2+2x-19$

And $g(x)=x+2$

To find the remainder, we use the Remainder Theorem, which states that when f(x) is divided by (x-c) then the f(c) is the required remainder.

So,

Here, we have,

$x+2=0\\\\x=-2$

So, we will find f(-2) which will give us the required remainder,

$f(-2)=3(-2)^4+2(-2)^3-(-2)^2+2(-2)-19\\\\f(-2)=(3\times 16)-(2\times 8)- 4-4-19\\\\f(-2)=48-16-36-4-19\\\\f(-2)=5$

Hence, Second option is correct.

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2 years ago

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Step-by-step explanation:

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Fantom [35]

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Step-by-step explanation:

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