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musickatia [10]
3 years ago
13

Rosita can wax her car in 2 hours or 120 minutes. When she works together with Helga, they can wax the car in 45 minutes. How lo

ng would it take Helga, working by herself, to wax the car?
Mathematics
1 answer:
Mashcka [7]3 years ago
6 0
<h2>Time taken by Helga to wax the car is 72 minutes</h2>

Step-by-step explanation:

Let w be the work of waxing.

Rosita can wax her car in 2 hours or 120 minutes.

Time taken by Rosita = 120 minutes

\texttt{Rate of Rosita = }\frac{W}{120}

Let time taken by Helga be t,

\texttt{Rate of Helga = }\frac{W}{t}

When she works together with Helga, they can wax the car in 45 minutes.

We have

             \frac{W}{\frac{W}{120}+\frac{W}{t}}=45\\\\120t=45(120+t)\\\\120t=5400+45t\\\\t=72minutes

Time taken by Helga to wax the car is 72 minutes

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Great to Least

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How do you graph -1/5x-1 and 4/5x-6
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Answer:

The points for the given to linear equations is (5 , - 2) and (5 , - 1)

The points is plotted on the graph shown .

Step-by-step explanation:

Given as :

The two linear equation are

y = \dfrac{-1}{5}x - 1                  ...........1

y = \dfrac{4}{5}x - 6                  ...........2

Now, Solving both the linear equations

Put the value of y from eq 2 into eq 1

I.e  \dfrac{4}{5}x - 6 = \dfrac{-1}{5}x - 1

Or, \dfrac{4}{5}x + \dfrac{1}{5}x  = 6 - 1

Or,  \dfrac{4 + 1}{5}x = 5

or, \dfrac{5}{5}x = 5

∴ x = 5

Now, Put the value of x in eq 1

So, y = \dfrac{-1}{5}x - 1      

Or, y = \dfrac{-1}{5}× 5 - 1              

or,  y = \dfrac{-5}{5} - 1

Or, y = - 1 - 1

I.e y = -2

So, For x = 5 , y = - 2

Point is (x_1 , y_1) = (5 , - 2)

Again , put the value of x in eq 2

So, y = \dfrac{4}{5}x - 6

Or, y = \dfrac{4}{5}× 5 - 6

Or, y = \frac{4\times 5}{5} - 6

Or, y = 4 - 6

I.e y = - 2

So, For x = 5 , y = - 2

Point is (x_2 , y_2) = (5 , - 2)

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The points is plotted on the graph shown . Answer

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3 years ago
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Answer:

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Step-by-step explanation:

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111=1(2)^2+1(2)^1+1(2)^0

So 1011+111 gives us:

1(2)^3+0(2)^2+1(2)^1+1(2)^0

+

1(2)^2+1(2)^1+1(2)^0

-----------------------------------------------------

Combine like terms:

1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0

1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace 2^0 with 1 and 2 with (2)^1:

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I want a 2^0 number:

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Combine like terms:

1(2)^3+2(2)^2+1(2)^1+0(2)^0

2(2)^2=2^3:

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We can rewrite the first term by law of exponents:

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So the binary form is:

10010

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

      1     0     1      1

+            1      1      1

------------------------------

     (1)    (1)    (1)

      1     0     1      1

+            1      1      1

------------------------------

     1 0    0     1       0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

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Then, if the two spinners are rotated, the sample space would be:

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The tree diagram shown in the attached image shows the 15 possible results

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{G, 2} {G, 3}

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3 years ago
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