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bulgar [2K]
3 years ago
6

What do solving absolute value equations and solving quadratic equations have in common?

Mathematics
2 answers:
NISA [10]3 years ago
6 0

Answer:

Step-by-step explanation:

Quadratic equations are equations that when rearranged using the 5 operations can yield a polynomial of degree 2 on one side of the equation and 0 on the other. There are 3 commonly taught methods for solving quadratic equations. These methods are factoring, completing the square, and the quadratic formula. These functions are most commonly used to model projectiles, disregarding air resistance.

When using the factoring method one attempts to write a degree 2 polynomial as the product of two degree 1 polynomials. This method works because if we multiply two degree one polynomials together we find  

For example: When factoring , we want to find two number a, b whose product is 6 and sum to 5. The values of a and b we are looking for are 2 and 3.

The method of completing the square centers around using the five operations to replace the original equation with an equivalent equation of the form  by taking advantage of the fact that

The absolute value function takes the value of a number, regardless of whether it is positive or negative. Some problems that the absolute value is useful for modeling include an object bouncing on the ground.

For example, , since we do not care about the negative sign.

Example: Solve  

Solution: Since the absolute value does not care about whether a number is positive or negative, if  then  or . Then we can solve each equation individually to find that x = 1 or 5.

Dafna1 [17]3 years ago
5 0

Answer:

They often have more than one solution.

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If 3 dogs were 2 ft and 4 were 5 what would be the total height of all of them together
seraphim [82]

Answer:

26 i think

Step-by-step explanation:

since the 3 dogs are 2 feet tall you would multiply them then multiply 4 by 5 and get 20 and 3x2 is 6 you add it up and get 26

7 0
3 years ago
Read 2 more answers
Help me pls. Pls do this step by step. I will give he brainliest to the person who does it step by step.
Zanzabum

Answer:

x^2 - x^3 - 3x^2y - 3xy^2

Step-by-step explanation:

x^3 +y^3 - (x + y)^3​

Expand the expression

x^2 + y^3 - (x^3 + 3x^2y + 3xy^2 +y3)

Remove the parentheses

x^2 +y^3 -x^3 -3x^2y -3xy^2 - y^3

Remove the opposites

Answer:

x^2 - x^3 - 3x^2y - 3xy^2

Hope this Helps!

3 0
2 years ago
Consider the set whose elements are the graphs having vertex set {1, 2, 3, 4}, and consider the relation on that set, where two
Damm [24]

Answer:

7

Step-by-step explanation:

Let S be the set of all graphs having vertex set  \{1,2,3,4\}. The relation \rho is defined over S such that

the graphs G and H are equivalent provided that they have same number of edges. Then, the number of equivalence classes depends on how many edges can be there in the vertex set \{1,2,3,4\} .

The number of edges is 0 forms a disconnected graph which makes an equivalent class.

The graphs of 1 edge makes an equivalent class.

The graphs of 2 edges makes an equivalent class.

The graphs of 3 edges makes an equivalent class.

The graphs of 4 edges makes an equivalent class.

The graphs of 5 edges makes an equivalent class.

In similar way, the only graph of 6 edges is complete graph which forms another equivalent class.

Hence,the total number of equivalent classes is 7.

8 0
2 years ago
Each swing that Susan builds requires $10 in hardware and $20 in wood. She uses the expression 10 + 20 to represent the total co
jenyasd209 [6]
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7 0
3 years ago
Please explain me how to solve it <br> cos2x+cos4x ≥0
Sliva [168]
2x + 4x will always be larger or equal to 0.
if x= 0 than it will be equal to 0.
Does this help at all?

3 0
3 years ago
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