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marin [14]
2 years ago
12

X(4-k)=p solve for k

Mathematics
2 answers:
kirill [66]2 years ago
6 0

Answer:

k=4-p\frac{p}{x}

Step-by-step explanation:

Given : Expression x(4-k)=p

To find : Solve for k?

Solution :

Step 1 - Write the expression,

x(4-k)=p

Step 2 - Divide by x both side,

\frac{x(4-k)}{x}=\frac{p}{x}

4-k=p\frac{p}{x}

Step 3 - Taking k to one side,

k=4-p\frac{p}{x}

Therefore, k=4-p\frac{p}{x}

den301095 [7]2 years ago
4 0
<span> x(4-k)=p
[x(4-k)=p]/x

4-k=p/x
-4 = -4
-k=p/x-4
k=4-p/x
</span>
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On the first day a total of 40 items were sold for $356. Define the variables and write a system of equations to find the number
Alina [70]
<h3><u><em>Question:</em></u></h3>

On the first day, a total of 40 items were sold for $356. Pies cost $10 and cakes cost $8. Define the variables, write a system of equations to find the number of cakes and pies sold, and state how many pies were sold.

<h3><em><u>Answer:</u></em></h3>

The variables are defined as:

"c" represent the number of cakes sold and "p" represent the number of pies sold

The system of equations used are:

c + p = 40 and 8c + 10p = 356

18 pies and 22 cakes were sold

<h3><em><u>Solution:</u></em></h3>

Let "c" represent the number of cakes sold

Let "p" represent the number of pies sold

Cost of 1 pie = $ 10

Cost of 1 cake = $ 8

Given that total of 40 items were sold

number of cakes + number of pies = 40

c + p = 40 ------ eqn 1

<u><em>Given items were sold for $356</em></u>

number of cakes sold x Cost of 1 cake + number of pies sold x Cost of 1 cake = 356

c \times 8 + p \times 10 = 356

8c + 10p = 356  ----- eqn 2

<u><em>Let us solve eqn 1 and eqn 2</em></u>

From eqn 1,

p = 40 - c    ---- eqn 3

Substitute eqn 3 in eqn 2

8c + 10(40 - c) = 356

8c + 400 - 10c = 356

-2c = - 44

c = 22

<em>Substitute c = 22 in eqn 3</em>

p = 40 - c

p = 40 - 22

p = 18

Thus 18 pies and 22 cakes were sold

3 0
2 years ago
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