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3 years ago

Help i need help i cant solve this

Mathematics

2 answers:

Hitman42 [59]3 years ago

7 0

To solve this equation, we must get the variable g alone on the left side of the equation. To do this, we need to “undo” the division on the left side of the equation by multiplying both sides by 4.

g/4 = 3.2

g/4 * 4 = 3.2 * 4

g = 12.8

Therefore, your answer is g = 12.8.

Hope this helps!

irakobra [83]3 years ago

3 0

Multiply both sides of the equation by 4 ( so g/4 becomes g):-

g/4 * 4 = 3.2 * 4

g = 12.8 (answer)

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2 years ago

Subtract the expressions.

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Answer:

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2 years ago

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2 years ago

Read 2 more answers

During the Playoff the schools football team has on average 30% their games than during the regular season. The average for this

exis [7]

Answer:12000

Step-by-step explanation:

During the Playoff the schools football team has on average 30% their games than during the regular season

The average for this year's playoffs attendance was 15,600 fans

To get the number of fans during a regular season goes thus:

Since there is an increase of 30%, we add 100% to 30% which gives:

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2 years ago

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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 $x = -1$

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3 years ago