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maw [93]
3 years ago
5

Number 6 part a & b Number 7 Please & thank you

Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

6- Part A: 750a+1200r=1000000

    Part B: 1065a+876r=1000000

Solution of the system: a\approx521.739\;\;\;\;b\approx507.246

Step-by-step explanation:

<u>Part A:</u>

In words, the equation is formed like this:

Total kg of Arabica + Total kg of Robusta = Total kg of coffee

Now, the total of each yield is calculated by the product of kg/hectare and number of hectares. Thus, the equation is:

750a+1200r=1000000

<u>Part B:</u>

In words:

Arabica's worth + Robusta's worth  = plantation worth

Each yield's worth is given by the product of the price per kg and the corresponding production. So, the equation is:

1.42(750a)+0.73(1200r)=1000000\\1065a+876r=1000000

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Each small square on the grid is 1 cm².
taurus [48]

Answer: The correct option is option third, i.e.,25 cm square.

Explanation:

From the given figure it is easily noticed that the shaded region contains two congruent parallelograms and 4 congruent triangles.

Area of a parallelogram is the product of base and height.

\text{Area of a parallelogram}=4\times 3=12

The area of a parallelogram is 12 cm square.

Area of a triangle is half of the product of base and height.

\text{Area of a triangle}=\frac{1}{2}\times 2\times 0.25=0.25

The area of a triangle is 0.25 cm square.

Since there are 2 parallelograms and 4 triangles in the figure, therefore the total area is,

A=2(12)+4(0.25)

A=24+1=25

Therefore, the area of figure is 25 cm square and the third option is correct.

4 0
3 years ago
Read 2 more answers
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).

So

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

(-\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).

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(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.

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3 years ago
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Answer:

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Step-by-step explanation:

Really, its just because, that's my explanation.

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