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ivolga24 [154]
3 years ago
6

Every six customer at flower shop receives a free rose in every ninth customer see is a free lily what customer will be the firs

t to receive a free rose and a lily
Mathematics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

18th customer

Step-by-step explanation:

The computation is shown below

Here we have to write the multiples of 6 and multiples of 9 that are presented below:

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48

Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72

So based on the above multiples of 6 and 9, 18th customer would first to receive a free rose and a lily

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Please answer correctly !!!!!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!
Over [174]

Answer:

n = 11

General Formulas and Concepts:

  • Order of Operations: BPEMDAS
  • Equality Properties
  • Vertical Angles: Angles that are across from each other and are congruent

Step-by-step explanation:

<u>Step 1: Set up equation</u>

<em>Vertical angles must be congruent.</em>

(6n - 4)° = (5n + 7)°

<u>Step 2: Solve for </u><em><u>n</u></em>

  1. Subtract 5n on both sides:                    n - 4 = 7
  2. Add 4 to both sides:                               n = 11
8 0
3 years ago
6x-5y=15 x=y+3 find x and y
arlik [135]

Answer:

(0, -3)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Coordinates (x, y)
  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

6x - 5y = 15

x = y + 3

<u>Step 2: Solve for </u><em><u>y</u></em>

<em>Substitution</em>

  1. Substitute in <em>x</em>:                                                                                                  6(y + 3) - 5y = 15
  2. Distribute 6:                                                                                                    6y + 18 - 5y = 15
  3. Combine like terms:                                                                                        y + 18 = 15
  4. [Subtraction Property of Equality] Subtract 18 on both sides:                     y = -3

<u>Step 3: Solve for </u><em><u>x</u></em>

  1. Define original equation:                                                                               x = y + 3
  2. Substitute in <em>y</em>:                                                                                                   x = -3 + 3
  3. Add:                                                                                                                   x = 0
7 0
3 years ago
Please help me with this homework
dimulka [17.4K]

Answer:

12

Step-by-step explanation:

49 - 37 = 12

or

It takes 12 jumps to get from 37 (a) to 49 (c)

8 0
3 years ago
Tom exercised 4/5 hour on monday and 5/6 hour on tuesday. Complete the calculations below to write equivalent fractions with a c
stepladder [879]
4/5= 24/30 5/6= 25/30 Equals in all= 1 19/30
7 0
3 years ago
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

7 0
2 years ago
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