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a_sh-v [17]
4 years ago
11

Can somebody please help me with number 20?!

Mathematics
2 answers:
Firdavs [7]4 years ago
8 0
It is easily b because you have to subtract the demensions and you get 8 squared
anyanavicka [17]4 years ago
8 0
Answer: Choice B, 8 square cm

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

Explanation:

A1 = shaded area inside larger rectangle 
A2 = shaded area inside smaller rectangle
A3 = unshaded overlapping region

A4 = A1+A2 represents the area of the entire 6 by 8 rectangle
A4 = 8*6 
A4 = 48

Similarly,
A5 = A2+A3 is the area of the entire 4 by 5 rectangle
A5 = 4*5 
A5 = 20

So we know that
A4 = A1+A3 = 48
A5 = A2+A3 = 20

Or in short,
A1+A3 = 48
A2+A3 = 20

Subtract the equations
A1+A3 = 48
A2+A3 = 20
----------
A1-A2 = 28

The total shaded region is 52, which means,
A1+A2 = 52

The new system of equations is
A1-A2 = 28
A1+A2 = 52
which combine to 2A1 = 80 and that solves to A1 = 40

If A1 = 40, then A2 = 12 after using A1+A2 = 52

Then we use A1+A3 = 48 to find that A3 = 8


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anyanavicka [17]

Answer:

Step-by-step explanation:

-8s-28>4

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-8s>32 we divide by 8

-s>4 we multiply by -1 and we will change the sign because of this

s<-4

4 0
3 years ago
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5/7 x 14/20<br><br> Snannan
bearhunter [10]

Answer:

\frac{1}{2}  \: or \: 0.5

Step-by-step explanation:

\frac{5}{7}  \times  \frac{14}{20}  =  \frac{5 \times 14}{7 \times 20}

It can be also written as -

=  >  \frac{5 \times 14}{20 \times 7}  =  \frac{5}{20}  \times  \frac{14}{7}  =  \frac{1}{4}  \times 2 =  \frac{1}{2}

7 0
3 years ago
A company finds that if they price their product at $ 35, they can sell 225 items of it. For every dollar increase in the price,
Basile [38]

Answer:

Maximum revenue = $8000

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Step-by-step explanation:

Given that:

Price of product = $35

Total sale of items = 225

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The total cost of item sold = 225 ×35

The total cost of item sold = 7875

If c should be the dollar unit in price increment;

Therefore; the cost function is : [35+c(1)][225-5(c)]

For maximum revenue;

\dfrac{d}{dc}(cost \ function) =0

\dfrac{d}{dc}[[35+c(1)][225-5(c)]]=0

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225 - 175  =10c

50 = 10c

c = 50/10

c = 5

Maximum revenue = [35+c(1)][225-5(c)]

Maximum revenue = [35+5(1)][225-5(5)]

Maximum revenue = (35 + 5)(225-25)

Maximum revenue = (40 )(200)

Maximum revenue = $8000

The price that will guarantee the maximum revenue is :

=(35 +c)

= 35 + 5

= $40

8 0
3 years ago
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When you watch this from a graphical point of view, the range is the portion of the y axis that the graph spans.

In this case, you can see that the lowest point reached by the function (its minimum) is (3,-2). You can also see that the function is unbounded upwards.

So, the range is [2,\infty), and the lowest value is thus 2.

7 0
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==-0['=98[=-6098979468039850-245[]7p76][5p6=??
uysha [10]

Answer:

My brain hurt

Step-by-step explanation:

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