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Anna71 [15]
3 years ago
9

Evaluate - cd/e when c = -9, d = -20, and e = -12. P.S. it’s not multiple choice.

Mathematics
2 answers:
Luda [366]3 years ago
6 0

Answer:

\boxed{ \frac{cd}{e} = - 15}

Given:

c =  - 9 \\ d =  - 20 \\ e =  - 12

Step-by-step explanation:

=  >  \frac{cd}{e}  \\  \\  =  >  \frac{( - 9)( - 20)}{( - 12)}  \\  \\  =  >  \frac{9 \times 20}{( - 12)}  \\  \\  =  >  -  \frac{180}{12}  \\  \\  =  >  - 15

yKpoI14uk [10]3 years ago
3 0

Answer:

-9*-20/-12=-15

Step-by-step explanation:

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What is the y-intercept of y = -3x
vlabodo [156]

Answer:

X=0

Step-by-step explanation:

8 0
3 years ago
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
zubka84 [21]

Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of​ butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

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?

From the question, we are told that:

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

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most ​$144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of​ butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9​% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

3 0
3 years ago
(16 \8) + (24 \ 6)<br> 2<br> DONE
Marina86 [1]

Answer:

6

Step-by-step explanation:

Remember the Order of Operations:

Parentheses

Exponents

Multiplication/Division (whichever comes first)

Addition/Subtraction (whichever comes first)

In this case, we start by simplifying what is in the parentheses:

(16÷8)+(24÷6)

2+4

6

3 0
2 years ago
Read 2 more answers
Please help me answer these questions
Triss [41]
Discussed why 2 is the answer
5 0
3 years ago
The diagram shows a solid shape.
Stolb23 [73]

Answer:

756\pi

Solution,

Volume of cone=270 pi cm^3

\frac{1}{3} \pi \:  {r}^{2}  = 270\pi \\ or \:  {r}^{2}  \times 10 =   270 \times 3 \\ or \:  {r}^{2}  \times 10 = 810 \\ or \:  {r}^{2}  =  \frac{810}{10}  \\ or \:  {r}^{2}  = 81 \\ or \: r =  \sqrt{81}  \\ or \: r =  \sqrt{ {9}^{2} }  \\ r = 9 \: cm

Volume of semisphere:

\frac{2}{3} \pi {r}^{3}  \\  = \:  \frac{2}{3}  \times \pi \times 9 \times 9 \times 9 \\  = 486\pi

Total:

270\pi + 486\pi \\  = 756\pi

hope this helps...

Good luck on your assignment..

6 0
2 years ago
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