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barxatty [35]
3 years ago
11

-3(2n-5)=1/2(-6n+30)

Mathematics
1 answer:
Anna71 [15]3 years ago
4 0
N=0
hope this helps jkkiuhbbjjj
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Please help me and ill mark you brainlest
dedylja [7]

Answer:

D

Step-by-step explanation:

Multiply by 3

4 x 3 = 12

9 x 3 = 27

8 0
2 years ago
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A large patio brick weighs 4 3/8 pounds. A small patio brick weighs 2 1/3 pounds. How much more does the large brick weigh?
Nikolay [14]

Subtracting the weight of the smaller brick from the larger gives us the weight difference:

4 3/8 lb - 2 1/3 lb. The two denominators are 8 and 3 respectively, resulting in an LCD of 24. Thus, our problem becomes:

4 9/24 - 2 8/24, which equals 2 1/24 lb. The weight difference is 2 1/24 lb.

5 0
3 years ago
6,657 ÷ 6 = Whats the answer and if there is a remander what is the remander
sdas [7]
The answer is 1,109.5
8 0
2 years ago
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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
Between which pair of decimals does the square root of 13 fall between
creativ13 [48]
Since 3 squared is 9 and 4 squared is 16, the square root of 13 will fall somewhere between 3 and 4
8 0
2 years ago
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