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san4es73 [151]
3 years ago
15

During the holidays, Duncan works as a gift wrapper. He works 35 hours a week, spread equally over 5 workdays. If Duncan earns $

9 an hour, how much does he make on each workday?
Mathematics
2 answers:
kenny6666 [7]3 years ago
5 0

Answer: $63

Step-by-step explanation: since 1h=$9

35 hours a week , works 5 days

35/5= 7 hours per day

Angelina_Jolie [31]3 years ago
3 0
Yes my mom and mom I got a baby text you baby text me when your baby girl love mom
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Solve x2 + 8x = 33 by completing the square. Which is the solution set of the equation? {–11, 3} {–3, 11} {–4, 4} {–7, 7}
NemiM [27]

Answer:

The solution of the equation are 3 , -11

Step-by-step explanation:

* Lets revise how to make the completing square

- The form of the completing square is a(x - h)² + k, where a , h , k

 are constant

- The general form of the quadratic is ax² + bx + c, where a , b , c

 are constant

- To change the general form to the completing square form equate

  them and find the constant a , h , k

* Now lets solve the problem

∵ x² + 8x = 33 ⇒ subtract 33 from both sides

∴ x² + 8x - 33 = 0

- lets change the general form to the completing square

∴ x² + 8x - 33 = a(x - h)² + k ⇒ solve the bracket of power 2

∴ x² + 8x - 33 = a(x² - 2hx + h²) + k ⇒ multiply the bracket by a

∴ x² + 8x - 33 = ax² - 2ahx + ah² + k ⇒ compare the two sides

∵ x² = ax² ⇒ ÷ x²

∴ a = 1  

∴ -2ah = 8 ⇒ substitute the value of a

∴ -2(1)h = 8 ⇒ -2h = 8 ⇒ ÷ (-2)

∴ h = -4

∵ ah² + k = -33 ⇒ substitute the value of a and h

∴ (1)(-4)² + k = -33

∴ 16 + k = -33 ⇒ subtract 16 from both sides

∴ k = -49

∴ x² + 8x - 33 = (x + 4)² - 49

* Now lets solve the completing square

∵ (x + 4)² - 49 = 0 ⇒ add 49 to both sides

∴ (x + 4)² = 49 ⇒ take square root for both sides

∴ (x + 4) = ± 7

∵ x + 4 = 7 ⇒ subtract 4 from both sides

∴ x = 3

∵ x + 4 = -7 ⇒ subtract 4 from both sides

∴ x = -11

* The solution of the equation are 3 , -11

4 0
3 years ago
Read 2 more answers
What is the inverse of the function <br><img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7Bx%20%2B%201%7D%7Bx%7D%20
expeople1 [14]
Y=(x+1)/x

x=y+1 / y

x=1/(y-1)
3 0
3 years ago
Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B
jeka94
\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0

M_y=35x^4+14x^5-6y^2-12xy^2
N_x=35x^4-6y^2

\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2

This suggests an integrating factor depending on x only is possible, and given by

\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}

Distributing across the ODE, we end up with

\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0

The equation is now exact, with

{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}

Now we find the solution:

F_x=M^*
F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx
F=(7x^5y-2xy^3)e^{2x}+f(y)

F_y=N^*
(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}
f'(y)=0
\implies f(y)=C

The general solution is then

F(x,y)=(7x^5y-2xy^3)e^{2x}=C
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4 years ago
1.) (2)(8+2i)= <br> Multiplying complex numbers
Nutka1998 [239]
Your answer is 16+4i
4 0
3 years ago
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What is the solution to -4(8 – 3x) &gt;_ 6x – 8?
Vsevolod [243]

Answer:

x ≥ 4

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

Step-by-step explanation:

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

-4(8 - 3x) ≥ 6x - 8

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

  1. Distribute -4:                              -32 + 12x ≥ 6x - 8
  2. Subtract 6x on both sides:        -32 + 6x ≥ -8
  3. Add 32 on both sides:              6x ≥ 24
  4. Divide 6 on both sides:             x ≥ 4

Here we see that <em>x</em> can be any value greater than or equal to 4.

3 0
3 years ago
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