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FinnZ [79.3K]
4 years ago
11

Question 1................

Mathematics
1 answer:
rewona [7]4 years ago
5 0
I think it is 9 that it ok
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4. Find the slope of the line through the points (-6, 4) and (8,-4). m =​
Aleksandr-060686 [28]

Answer: Hope it helps :)

<u>-4</u>

7

Step-by-step explanation:

You have to use the formula <u>y2-y1</u>

                                                x2-x1

So <u>-4-4</u>

     8--6

-4-4= -8

8--6= 14 (two negative signs become a positive sign)

You have -8 over 14

If you simplify it's -4/7 or <u>-4</u>

                                         7

4 0
3 years ago
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Pls help me answered this paper. I already answered A but I'm to lazy to answers the rest. SORRY ;-; lol
Vinil7 [7]
When looking for volume always multiply the sides to find the answer
5 0
4 years ago
Alison was late paying her credit card bill of $247. She was charged a 5% late fee. What was the amount (in dollars) of the late
irga5000 [103]

It is $12.35, because this number is 5% of 247.

8 0
4 years ago
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Melissa working alone could clean one third of the area in one hour how much of the area could she clean in half of an hour.
docker41 [41]
Melissa could clean one sixth of the area in half and hour because you divide by 2 since 60 divided by 2 is 30 and she has half as much time
5 0
3 years ago
2. Find the stationary points for the function
vovikov84 [41]

Answer & step-by-step explanation:

Stationary points are the points where the first derivative is equal to zero.

Let's calculate it using the power rule (exponent comes forward, decrease exponent by 1) and the fact that the derivative is a linear operation (that is D[a\ f(x) + b\ g(x)] = a Df(x) + b Dg(x))

The first derivative is then

y' = \frac13 (3x^2) - \frac12 (2x) -6 = x^2-x-6 = (x+2)(x-3)

Note that the last passage is not strictly needed, but it's really helpful to find stationary points, when in this next passage we set it equal to zero. Alternatively, you can use the quadratic formula if you can pull the factors out of your head right away.

y'=0 \rightarrow (x+2)(x-3) = 0 \implies x=-2  || x=3

These two point could be maxima, minima, or inflection points. To check them you can either see how the sign of the first derivative goes, or check the sign of the second derivative, as you're required.

The rules states that if the second derivative evaluated in that point is negative we have a maximum, if it's positive we have a minimum, and if we have a zero we keep derivating until we get a non-zero  value.

In our case, the second derivative we get by calculating the derivative again and we get y'' = 2x-1 . Evaluating it at both points we get

y''(-2) = 2(-2)-1 = -5\\y''(3) = 2(3) -1 = 5

so -2 is a maximum and 3 is a minimum.

5 0
2 years ago
Read 2 more answers
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