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2 years ago

Help please !!!!!!!!!!!!!!

Mathematics

1 answer:

DIA [1.3K]2 years ago

8 0

The LCM will be the values that are common to both factors. From the given values the LCM will be 2² * 3² * 5

<h3>Least common multiple</h3>

LCM is the lowest number that can divide all other numbers given in an expression.

Given the prime factorizations 2² * 3² * 5 and 2*3*5. The LCM will be the values that are common to both factors.

From the given values the LCM will be 2² * 3² * 5

Learn more on LCM here: brainly.com/question/233244

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3 years ago

In this bar how many groups of 80 make 60

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3 years ago

Janet is three years older than her sister julie. Janet’s brother is eight years younger than their sister Julie. The sum of all

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Answer:

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Step-by-step explanation:

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3 years ago

(1) Which of the following is the factored form of:<br> x2 + 10x +9

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Answer:

( + 1 ) ( + 9 )

Step-by-step explanation:

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3 years ago

Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.

liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

$f(x) = \frac{x^2+4x+3}{x^2-x-2}$

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial $x^2-x-2$

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

$(x-2) (x + 1)$

We do the same with the numerator

$x^2+4x+3$

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

$(x+3)(x + 1)$

Therefore

$f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}$

Note that $\frac{(x+1)}{(x+1)}=1$ only if $x \neq -1$

So since $x = -1$ is not included in the domain the function has a discontinuity in $x = -1$

3 0

3 years ago