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3 years ago

15

A salon charges $25 for a women haircut, $ 10 for a men haircut and $8 for a child haircut. The salon needs to make at least $43

00 a month. Write an inequality to represent the different hair cuts to reach or beat their goal.

1 answer:

topjm [15]3 years ago

5 0

Answer:

25w+10m+8c≥4300 (w: # of women, m: men, c: children)

Step-by-step explanation:

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PilotLPTM [1.2K]

Let's say Bailey takes "b" days to paint it.

and Andrew takes "a" days to paint the same house.

now, Andrew is 6 times faster than Bailey, therefore, if Andrew takes "a" days to do it, Bailey takes then "6a" days, or b = 6a.

now, the year they worked together, they finished it in 7 days.

so, after 1 day then, they have only done 1/7 of the whole work.

and Andrew for one day, has done 1/a of the house, whilst Bailey has done 1/b of the house or 1/(6a).

$\bf \stackrel{\textit{Andrew's rate}}{\cfrac{1}{a}}+\stackrel{\textit{Bailey's rate}}{\cfrac{1}{b}}=\stackrel{\textit{1 day of work}}{\cfrac{1}{7}}
\\\\\\
\cfrac{1}{a}+\cfrac{1}{6a}=\cfrac{1}{7}\impliedby 
\begin{array}{llll}
\textit{let's multiply all by }\stackrel{LCD}{42a}\textit{ to toss the}\\
denominators
\end{array}$

$\bf 42a\left( \cfrac{1}{a}+\cfrac{1}{6a} \right)=42a\left( \cfrac{1}{7} \right)\implies 42+7=6a\implies \cfrac{49}{6}=a
\\\\\\
\stackrel{days}{8\frac{1}{6}}=a
\\\\\\
\textit{how many days will it take Bailey then?}\quad b=6a
\\\\\\
b=6\cdot \cfrac{49}{6}\implies b=\stackrel{days}{49}$

6 0

3 years ago

Can someone help me with my algebra homework I have 30 minutes until it's due 6 questions my insta is bahdanz

posledela

Answer:

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7 0

2 years ago

WILL GIVE BRAINLIEST FASTTTTT

Stells [14]

Answer: (x−2)(x−2)

The middle number is -4 and the last number is 4.

Factoring means we want something like

(x+_)(x+_)

Which numbers go in the blanks?

We need two numbers that...

Add together to get -4

Multiply together to get 4

Can you think of the two numbers?

Try -2 and -2:

-2+-2 = -4

-2*-2 = 4

Fill in the blanks in

(x+_)(x+_)

with -2 and -2 to get...

(x-2)(x-2)

Answer:

(x−2)(x−2)

so the answer is D.

6 0

3 years ago

Determine the order pair and quadrant for each point:<br> A.<br> B.<br> C.

Taya2010 [7]

Answer:

A: (-4,1) QUAD 2

B: (-1, -3) QUAD 3

C: (2,5) QUAD 1

D: (5, -4) QUAD 4

Step-by-step explanation:

4 0

3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago