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jonny [76]
3 years ago
6

Which integer did he write?

Mathematics
2 answers:
cluponka [151]3 years ago
7 0
-3 is the integer Phillip wrote
muminat3 years ago
5 0
The answer should be -3
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Write an expression for “the quotient for y and 8”
jonny [76]

Answer:


Step-by-step explanation:

I think your answer is y/8. It's a little unusually worded. If this is not correct, please leave a note.

The way I have answered it, I would word it as "What is the quotient of y divided by 8."


7 0
3 years ago
What proportion of a normal distribution corresponds to z scores greater than 1.04?
Tom [10]
We would need to look over the z table to find the area under the standard normal distribution curve to the left of z = 1.04. Then we'll subtract it from 1 to get the proportion of a normal distribution corresponding to z scores greater than 1.04.
By looking at the z table, we can see that the area to the left of z = 1.04 is 0.8508. So the proportion of a normal distribution to the right of z = 1.04 is 1 – 0.8508 =  0.1492.

The answer is 0.1492.
7 0
3 years ago
Read 2 more answers
X-1) x-x-2x+8<br> Long division
Nastasia [14]

Answer:

Ummm

Step-by-step explanation:

Im confused

4 0
3 years ago
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
The cosine of the complimentary angle to 58 degrees
lana [24]
Yes
Divide it by sixty and multiply
I hope this helps
5 0
3 years ago
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