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Digiron [165]
3 years ago
5

Please help! asap! thank you so much! I have photos attached!

Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:


Step-by-step explanation:

Given problems are absolute value problems, So we need to plug the values of given parameters and get the final result.

We have given here,

a =-2 , b = 3 ,  c = -4 and d = -6

Now we know that An absolute function always gives a positive value.

Let's apply this strategy in the given problems.

1. ║a+b║

Plug a= -2 and b = 3

We get, ║-2+3║=║1║= 1

2. 5║c+b║

Plug c= -4 and b=3

i.e.   5║-4 + 3║= 5║-1║=5×1 = 5

3. a+b║c║

Plug values a= -2 , b=3 and c=-4

i.e -2 +3║-4║ = -2 + 3×4 = -2 + 12 = 10

4. ║a+c║÷(-d)

i.e ║-2 + (-4)║÷(-6) = ║-6║÷(-6) = 6÷(-6) = -1

5. 3║a+d║+b

i.e 3║-2+(-6)║+3 = 3║-8║+3 = 3×8 +3 = 27

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A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co
lisov135 [29]

Answer:

The complete solution and step by step explanation is written below.

<em>Note:</em>

Here is the complete question you were probability asking as you may have unintentionally missed to mention the parts (a) and (b). So, after a little research, I figured out that this was probably the complete question you had meant to ask. So, I am answering based on the reference of this complete question that may indeed clear your concepts.

Complete Question:

<em>A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I costs ​$2.40 per​ gallon, and milk from dairy II costs ​$0.80 per gallon. At most ​$144 is available for purchasing milk. Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat, and dairy II can supply at most 80 gallons averaging 2.9​% butterfat. Answer parts a and b</em>.

<em>A. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of butterfat? (Also, what IS the maximum amount of butterfat?)</em>

<em>B. The solution from part A leaves both dairy 1 and dairy 2 with excess capacity. Calculate the amount of additional milk each dairy could produce.</em>

Step-by-step explanation:

Let A be number of gallons of the amount of milk the company bought from Dairy 1.

(100 - X) be the number of gallons of the amount of milk the company bought from Dairy 2.

Milk from dairy I costs ​$2.40 per​ gallon and milk from dairy II costs ​$0.80 per gallon.

So, the equation of total cost available for purchasing the milk be:

                            Cost =  2.40A + ( 100 - A)0.80

As $144 is the total cost available for purchasing the milk.

Hence, Cost equation would be as follows:

                             Cost =  2.40A + ( 100 - A)0.80

                               144 =  2.40A + ( 100 - A)0.80

                               144 = 2.40A + 80 - 0.80A

                              1.6A = 64

                            So, A = 64 ÷ 1.6

                                      = 40 gallons

<em>Calculation of gallons of Dairy 1 and Dairy 2:</em>

gallons of Dairy 1 ⇒ A = 40 gallons

gallons of Dairy 1 ⇒ 100 - A = 100 - 40 = 60 gallons

<em>The equation for Fat will be:</em>

Fat for dairy 1 = 40 × 3.9​% = 1.56

Fat for dairy 2 = 60 × 2.9​% = 1.74

ADD = 1.56 + 1.74 = 3.3 %

<em>Calculation Part B:</em>

Dairy 1 will have = 50 - 40 = 10 gallons ( excess)

Dairy 2 will have = 80 - 60 = 20 gallons (excess)

   

Keywords: word problem, equation solving

Learn more about word problems and equation solving from brainly.com/question/8917703

#learnwithBrainly

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A rectangular parking lot has a perimeter of 820 ft. The area of the parking lot measure SL 42,000 ft
horsena [70]

Answer:

a= 200

b = 210

Step-by-step explanation:

My assumption is, we have to find the length of sides of rectangle

Given

perimeter = 2a + 2b = 820 ft (i) (here a is smaller side and b is larger side)

area = a*b = 42,000 ft^2 (ii)

from eq (1)

2a + 2b = 820

=> 2(a+b) = 820

=> a+b = 820/2

=> a + b = 410

=> a = 410-b   (iii)

putting the value of a in eq(ii), we get

(410-b) *b = 42,000

410b - b^2 = 42,000

0 = b^2 - 410b + 42000

b^2 - 410b + 42000 = 0

b^2- 200b- 210b + 42000 = 0

b(b-200)-210(b-200) = 0

(b-200)(b-210) = 0

or

b= 210 and b = 200

if b is larger side than b =210

By putting value of b in eq(iii),

a = 410 -210 = 200

 

6 0
3 years ago
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