Solving systems of linear inequalities

Mathematics

1 answer:

Liono4ka [1.6K]4 years ago

4 0

Answer:

To solve a system of of linear inequalities, you could do the elimination method or substitution method.

Step-by-step explanation:

For elimination method, you can add the same value to each side of an equation. So if you have a system: x – 6 = −6 and x + y = 8, you can add x + y to the left side of the first equation and add 8 to the right side of the equation.

For substitution method, you have to:

Step 1 : First, solve one linear equation for y in terms of x.

Step 2 : Then substitute that expression for y in the other linear equation.

Step 3 : Solve this, and you have the x -coordinate of the intersection.

Step 4 : Then plug in x to either equation to find the corresponding y -coordinate.

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What is the answer of the equation<br> -0.35+(-0.75+0.4)

Evgen [1.6K]

In the order of operations, parentheses come before addition.

-0.75 + 0.4 = -0.35

-0.35 + -0.35 = -0.7.

-0.35 + (-0.75 + 0.4) = -0.7

Hope this helps! Have an awesome day :)

8 0

3 years ago

Um litro de suco é dividido entre três amigos. O primeiro fica com 14 do todo, o segundo fica com 1/3 do total e o terceiro fica

kkurt [141]

Answer:

a) 1/24 do total

Step-by-step explanation:

We have that:

The total is 1.

The first friend gets 1/4 of the total.

The second friend gets 1/3 of the total

The third friend gets 3/8 of the total.

The remaining is x.

$\frac{1}{4} + \frac{1}{3} + \frac{3}{8} + x = 1$

The least common multiple between 4, 3 and 8 is 24. So

$\frac{6 + 8 + 3*3 + 24x}{24} = 1$

$23 + 24x = 24$

$24x = 1$

$x = \frac{1}{24}$

So the correct answer is:

a) 1/24 do total

3 0

4 years ago

If x2+1/x2=18, find the value of x3−1/x3

Elden [556K]

Answer:

There are 2 answers : 76 or -76

Step-by-step explanation:

$x^2+\dfrac{1}{x^2} =18\\\\(x-\dfrac{1}{x} )^2=x^2+\dfrac{1}{x^2} -2=18-2=16\\\\x-\dfrac{1}{x}=4\ or\ x-\dfrac{1}{x}-4\\\\(x-\dfrac{1}{x} )^3=x^3-\dfrac{1}{x^3} -3x^2*\dfrac{1}{x}+3x*\dfrac{1}{x^2}\\\\=x^3-\dfrac{1}{x^3}-3(x-\dfrac{1}{x})\\\\x^3-\dfrac{1}{x^3}=(x-\dfrac{1}{x} )^3+3(x-\dfrac{1}{x})\\\\if\ x-\dfrac{1}{x}=4\ then \ x^3-\dfrac{1}{x^3}=4^3+3*4=64-12=76\\\\if\ x-\dfrac{1}{x}=-4\ then \ x^3-\dfrac{1}{x^3}=(-4)^3+3*(-4)=-64-12=-76\\\\$