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3 years ago

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Martha is a salon owner. Yesterday, she did 5 haircuts and colored the hair of 2 clients, charging a total of $421. Today, she d

id 1 haircut and colored the hair of 2 clients, charging a total of $233. How much does Martha charge for her services? Martha charges $_______ for a haircut and $_________ for a coloring.

1 answer:

Vanyuwa [196]3 years ago

4 0

Answer:

I'm not sure if I got this correct but here's my answer. Martha charges $60.14 for a haircut and she charges $77.66 for hair coloring.

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Line e passes through points (10, 8) and (3, 5). Line f passes through points (3, 1) and (10, 4). Are line e and f parallel or p

babunello [35]

Answer:

e║f.

Step-by-step explanation:

1) the pointed vectors of the given lines are:

for line e: (3-10;5-8)=(-7;-3);

for line f: (10-3;4-1)=(7;3);

2) -7/7=-3/3=-1, it means the lines are parallel.

3 0

2 years ago

Help me i need to submit to my teacher by 8pm

Vladimir [108]

Please see the figure. We'll first work out half the area of the rounded triangle, half the unshaded part, then double it, then subtract it from the big square.

Half the area is the circular sector PTQ (with center P, arc TQ) minus the right triangle PUT.

A/2 = area(sector PTQ) - area(triangle PUT)

The triangle is half of equilateral triangle PQT, so a 30/60/90 right triangle so we know the sides are in ratio 1:√3:2 so

TU = (7/2)√3

area(PUT) = (1/2) (7/2)(7/2)√3 = (49/8)√3

area(sector PTQ) = (angle TQP / 360°) πr^2

We know angle TQP is 60° because TQP is equilateral. r=7.

area(sector PTQ) = (60°/360°) π (7²) = 49π/6

Putting it together,

A/2 = area(sector PTQ) - area(triangle PUT)

A = 2(49π/6 - (49/8)√3)

A = 49(π/3 - √3/4) square cm

I hate ruining a nice exact answer with an approximation, but they seem to be asking.

A ≈ 30.095057615914535

Check:

I'm not sure how to check it. I'd estimate it's about 25% bigger than equilateral triangle PQT with area (√3/4)7² ≈ 21.2, so around 27. 30 seems reasonable.

Now the real area we seek is the big square PQRS minus A, so

area = 7² - 30.095057615914535 = 18.904942384086 sq cm

They want square meters for some reason; we scale by (1/100)²

Answer: 0.00189 square meters

7 0

3 years ago

100%<br> &<br> There are 16 girls and 18 boys in a class. The teacher chooses a student's name at<br> What is the probabilit

Mariulka [41]

Answer:a

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

the ratio of sadia's age to her father's age is 3:6. The sum of their age is 96 .What is sadia's age

max2010maxim [7]

We have,

$a:b=3:6,a+b=96$

Introduce variable $x$ such that $a=3x,b=6x$

The sum $a+b=96$ is therefore $9x=96\implies x=10.\overline{6}$

So,

$a=3\cdot10.\overline{6}=\boxed{32}$ (sadia's age)

$b=6\cdot10.\overline{6}=\boxed{64}$ (father's age)

Hope this helps :)

4 0

3 years ago

The graph of f is given in the figure to the right. Let A(x)equals=Integral from 0 to x f left parenthesis t right parenthesis

Tpy6a [65]

Answer:

$A(4)=-4\pi$

$A(8)=-4\pi +8$

$A(12)=-4\pi +16$

$A(14)=-4\pi +15$

Step-by-step explanation:

we are given

$A(x)=\int\limits^x_0 f{x} \, dx$

Calculation of A(4):

we can plug x=4

$A(4)=\int\limits^4_0 f{x} \, dx$

Since, this curve is below x-axis

so, the value of integral must be negative

and it is quarter of circle

so, we can find area of quarter circle

radius =4

$A(4)=-\frac{1}{4}\times \pi \times (4)^2$

$A(4)=-4\pi$

Calculation of A(8):

we can plug x=8

$A(8)=\int\limits^8_0 f{x} \, dx$

we can break into two parts

$A(8)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx$

now, we can find area and then combine them

$A(8)=-4\pi +\frac{1}{2}\times 4\times 4$

$A(8)=-4\pi +8$

Calculation of A(12):

we can plug x=12

$A(12)=\int\limits^12_0 f{x} \, dx$

we can break into two parts

$A(12)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx$

now, we can find area and then combine them

$A(12)=-4\pi +\frac{1}{2}\times 8\times 4$

$A(12)=-4\pi +16$

Calculation of A(14):

we can plug x=14

$A(14)=\int\limits^14_0 f{x} \, dx$

we can break into two parts

$A(14)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx+\int\limits^14_12 f{x} \, dx$

now, we can find area and then combine them

$A(14)=-4\pi +\frac{1}{2}\times 8\times 4-\frac{1}{2}\times 1\times 2$

$A(14)=-4\pi +16-1$

$A(14)=-4\pi +15$

8 0

3 years ago