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Nata [24]
3 years ago
13

Which expression is equivalent to 1/4 (5x + 6)?

Mathematics
1 answer:
Anettt [7]3 years ago
4 0

The expression which is equivalent to 1/4 (5x + 6) is; Choice A: {5(1/4)x} + {6(1/4)x}.

According to the question:

  • We are required to determine an expression which is equivalent to <em>1/4 (5x + 6).</em>

In a bid to expand the expression; we must multiply each term in the parenthesis by (1/4);

In essence; we have;

  • <em>{5(1/4)x} + {6(1/4)x}</em>

Read more on multiplication;

brainly.com/question/847421

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The answer is 9x5-15
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a group raised $276 for its favorite charity. this is 23% of the amount it hopes to raise. how much money does the group want to
OverLord2011 [107]

Answer: The group wants to raise $1,200.

Step-by-step explanation:

We can use a proportion to solve.

part/whole = percent/100

If we insert the values:

276/x = 23/100

We can cross multiply.

276 x 100 = 27,600

27,600 / 23 = 1,200

$1,200 is the answer.

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Answer:

B. $10 bag, $0.80 per pound

Step-by-step explanation:

10 / 8 = 0.80

25 / 15 = 1.67

0.80 is cheaper

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How many groups of 1/2 day are in 1 week?
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Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
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