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BARSIC [14]
2 years ago
5

Write the sentence as an equation. the quotient of 331 and u is the same as 13

Mathematics
2 answers:
umka2103 [35]2 years ago
6 0
331 dived by u equals 13
snow_tiger [21]2 years ago
3 0
331(divide sign)u=13
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1/4( -1 + 4)+3/5 please help ME missing work!
nignag [31]

Answer:

1.35

Step-by-step explanation:

I believe this is right but it could be wrong. Hope this helps! <3

3 0
2 years ago
<img src="https://tex.z-dn.net/?f=%28%20%5Cfrac%7B1%7D%7B2%7D%29%5E%7B2%7D%20" id="TexFormula1" title="( \frac{1}{2})^{2} " alt=
Vladimir79 [104]
All you need to do is solve it
\frac{1}{2} ^{2} or \frac{1}{2}  x \frac{1}{2}
When solved your answer is 0.25 or 1/4
------------------------
What I did to solve it was....

1. Divide top fraction by bottom fraction to get decimal
1 ÷ 2 = 0.5

2. Then multiply 0.5 by its self (the same as putting it to the ^2)
0.5 x 0.5 = 0.25
5 0
3 years ago
A restaurant operator in Accra has found out that during the lockdown,if she sells a plate of her food for Ghc 20 each,she can s
ella [17]

The demand equation illustrates the price of an item and how it relates to the demand of the item.

  • The slope of the demand function is -1/2
  • The equation of the demand function is: R(x) = (300 - 10x) \times (20 + 5x)
  • The price that maximizes her revenue is: Ghc 85

From the question, we have:

Plates = 300

Price = 20

The number of plates (x) decreases by 10, while the price (y) increases by 5. The table of value is:

\begin{array}{cccccc}x & {300} & {290} & {280} & {270} & {260} \ \\ y & {20} & {25} & {30} & {35} & {40} \ \end{array}

The slope (m) is calculated using:

m = \frac{y_2 - y_1}{x_2 - x_1}

So, we have:

m = \frac{25-20}{290-300}

m = \frac{5}{-10}

m = -\frac{1}{2}

The equation of the demand is as follows:

The initial number of plates (300) decreases by 10 is represented as: (300 - 10x).

Similarly, the initial price (20) increases by 5 is represented as: (20 + 5x).

So, the demand equation is:

R(x) = (300 - 10x) \times (20 + 5x)

Open the brackets to calculate the maximum revenue

R(x) =6000 + 1500x - 200x - 50x^2

R(x) =6000 + 1300x - 50x^2

Equate to 0

6000 + 1300x - 50x^2 =0

Differentiate with respect to x

1300 - 100x =0

Collect like terms

100x =1300

Divide by 100

x =13

So, the price at maximum revenue is:

Price= 20 + 5x

Price= 20 + 5 * 13

Price= 85

In conclusion:

  • The slope of the demand function is -1/2
  • The equation of the demand function is: R(x) = (300 - 10x) \times (20 + 5x)
  • The price that maximizes her revenue is: Ghc 85

Read more about demand equations at:

brainly.com/question/21586143

3 0
3 years ago
Given the system of constraints name all vertices. Then find the maximum value of the given objective
larisa86 [58]

Answer:

35

Step-by-step explanation:

The constraints are

The red line represents the function

y\leq \dfrac{1}{3}x+1

At y=0

0=\dfrac{1}{3}x+1\\\Rightarrow -1=\dfrac{1}{3}x\\\Rightarrow x=-3

At x=0

y=0+1\\\Rightarrow y=1

Two points are (-3,0),(0,1)

The blue line represents the function

5\geq y+x

at y=0

5=x

at x=0

y=5

Two points are (5,0),(0,5)

The other two constraints are x\geq 0, y\geq 0. So, the point has to be in the first quadrant

From the graph it can be seen there are two points where the function will be maximum let us check them.

(3,2)

7x-2y=7\times 3-2\times 2=17

(5,0)

7x-2y=7\times 5-2\times 0=35

So, the maximum value of the function is 35.

3 0
2 years ago
Salvador’s class has collected 88 cans in a food drive. They plan to sort the cans into x bags, with an equal number of cans in
marusya05 [52]
I believe it’s C, I’m not every sure
7 0
2 years ago
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