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3 years ago

1/2x-2/3>3 enter your answer using interval notation

Mathematics

1 answer:

Nata [24]3 years ago

7 0

So basically u keep the x on one side and the rest on the other. Get a common denominator. to get rid of the half u times both sides by 2.
hope this helps

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Answer the following question about the function whose derivative is given below

Komok [63]

Answer:

a) The critical points are $x = 3$ and $x = -6$ .

b) f is decreasing in the interval $(-\infty, -6)$

f is increasing in the intervals $(-6,3)$ and $(3,\infty)$ .

c) Local minima: $x = -6$

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which $f^{\prime}(x) = 0$ . So

$f^{\prime}(x) = 0$

$(x-3)^{2}(x+6) = 0$

So, the critical points are $x = 3$ and $x = -6$ .

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if $f^{\prime}$ is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are $x = 3$ and $x = -6$ . So we have those following intervals:

$(-\infty, -6), (-6,3), (3, \infty)$

We select a point x in each interval, and calculate $f^{\prime}(x)$ .

-------------------------

$(-\infty, -6)$

$f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100$

f is decreasing in the interval $(-\infty, -6)$

---------------------------

$(-6,3)$

$f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8$

f is increasing in the interval $(-6,3)$ .

------------------------------

$(3, \infty)$

$f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10$

f is increasing in the interval $(3,\infty)$ .

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At $x = -6$ , f goes from decreasing to increasing.

So $x = -6$ , f assume a local minima value

At $x = 3$ , f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

4 0

3 years ago

Buns,cost 40p each. Cakes cost 55p each. I spend exactly £4.35 on buns and cakes.

horsena [70]

By solving a linear equation, we conclude that you bought 4 buns and 5 cakes.

We will define the two variables:

x = number of buns bought.
y = number of cakes bought.

We know that you spent exactly £4.35, then we only need to solve the linear equation:

x*0.40 + y*0.55 = 4.35

We need to solve that equation for x and y, such that the values of x and y can only be positive whole numbers.

We can rewrite the equation as:

y*0.55 = 4.35 - x*0.40

y = (4.35 - x*0.40)/0.55

Now we only need to evaluate it in different values of x, and see for which value of x, the outcome y is also a whole number.

We will see that for x = 4, we have:

y = (4.35 - 4*0.40)/0.55 = 5

So you bought 4 buns and 5 cakes.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0

2 years ago

Read 2 more answers

A bad contains 15 balls of which X is red. Write an expression for the probability that a ball drawn from the bad is red. When 5

antoniya [11.8K]

Answer:

Step-by-step explanation:

The current probability of picking a red ball is $\frac{x}{15}$ . When 5 more balls are added, the probability is 3/4. We can write a ratio for this problem: