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lesantik [10]
3 years ago
13

Can an equation with the variable on both sides be written as a two step equation

Mathematics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

·         Use properties of equality together to isolate variables and solve algebraic equations.

·         Use the properties of equality and the distributive property to solve equations containing parentheses, fractions, and/or decimals.

 

Introduction

 

There are some equations that you can solve in your head quickly. For example – what is the value of y in the equation 2y = 6? Chances are you didn’t need to get out a pencil and paper to calculate that y = 3. You only needed to do one thing to get the answer, divide 6 by 2.

 

Other equations are more complicated. Solving  without writing anything down is difficult! That’s because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.

 

Using Properties of Equalities

 

Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal.

 

If the equation is in the form, ax + b = c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.

 

 

Step-by-step explanation:

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3 years ago
What is absolute extrema of cube root of x on I=[-3,8]
hichkok12 [17]
<span>These are points where f ' = 0. Use the quiotent rule to find f '. 

f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2 
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2 

Set f ' (x) = 0 and solve for x. 

f ' (x) = 0 = (2-2x^3) / (x^3+2)^2 

Multiply both sides by (x^3+2)^2 

(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2] 
0 = 2 - 2x^3 

Add 2x^3 to both sides 

2x^3 + 0 = 2x^3 + 2 - 2x^3 
2x^3 = 2 

Divide both sides by 2 

2x^3 / 2 = 2 / 2 
x^3 = 1 

Take cube roots of both sides 

cube root (x^3) = cube root (1) 
x = 1. This is our critical point 

2) Points where f ' does not exist. 

We know f ' (x) = (2-2x^3) / (x^3+2)^2 

You cannot divide by 0 ever so f ' does not exist where the denominator equals 0 

(x^3 + 2)^2 = 0. Take square roots of both sides 
sqrt((x^3+2)^2) = sqrt(0) 
x^3 + 2 = 0. Add -2 to both sides. 
-2 + x^3 + 2 = -2 + 0 
x^3 = -2. Take cube roots of both sides. 
cube root (x^3) = cube root (-2) 
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point. 

3) End points of the domain. 

The domain was clearly stated as [0, 2]. The end points are 0 and 2. 

Therefore, our only options are: 0, 1, 2. 

Check the intervals 

[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign. 

In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive. 

In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative. 

Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum. 

f (0) = 0 
f (1) = 1/3 
f (2) = 1/5 

Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
3 0
4 years ago
Hello hello follow please ​
Tju [1.3M]

Answer:

ok

Step-by-step explanation:

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I need help on this asap
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Answer:

Option 4

Step-by-step explanation:

:)

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