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Verdich [7]
3 years ago
11

Allegra has a cell-phone plan that charges $65 per month and $0.10 for every minute that she uses the phone beyond what her plan

allows. One month, she was billed $73.40. Define a variable for the unknown. Write an equation to model the problem. Explain your answer. Solve the equation. Show your work. Find the number of minutes that Allegra went over the time that the plan allows. Explain your answer.
Mathematics
2 answers:
Sloan [31]3 years ago
8 0
65+0.1m=bill

bill=73.40

65+0.1m=73.4
minus 65 from both sides
0.1m=8.4
divide both sides by 0.1
m=84
went 84 mins over
Helga [31]3 years ago
3 0
X0.10. xtimesy (y being every minute) = 8.40 (Money that she spent extra) Well 8.40 divided by 0.10 is 84. so she went 84 mins over
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Adidea Corp. estimates that $5,670 of its accounts receivables are uncollectible. How will the company record the transaction?
faltersainse [42]

Answer:

When it is said that accounts receivables are uncollectible, it means that the company does not expect some of its accounts receivables to pay them back for the goods that the accounts receivables purchased on credit.

Uncollectible account receivables therefore reduce the accounts receivables account.

Entry will be:

Date              Accounts Description                                   Debit              Credit

xx-xx-xxxx     Noncollectible accounts expense             $5,670

                      Accounts Receivables                                                      $5,670

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8 0
3 years ago
Find the shaded area​
REY [17]

The parabola <em>y</em> = <em>x</em> ² and the line <em>x</em> + <em>y</em> = 12 intersect for

<em>x</em> ² = 12 - <em>x</em>

<em>x</em> ² + <em>x</em> - 12 = 0

(<em>x</em> - 3) (<em>x</em> + 4) = 0

===>   <em>x</em> = 3

so you can compute the area by using two integrals,

\displaystyle \int_0^3 x^2\,\mathrm dx + \int_3^{12}(12-x)\,\mathrm dx

Then the area you want is

\displaystyle \frac{x^3}3\bigg|_0^3 + \left(12x-\frac{x^2}2\right)\bigg|_3^{12} = \left(\frac{3^3}3-\frac{0^3}3\right) + \left(12^2-\frac{12^2}2 - 12\times3 + \frac{3^2}2\right) \\\\ = \boxed{\frac{99}2}

Alternatively, you can subtract the area bounded by <em>y</em> = <em>x</em> ², <em>x</em> + <em>y</em> = 12, and the <em>y</em>-axis in the first quadrant from the area of a triangle with height 12 (the <em>y</em>-intercept of the line) and length 12 (the <em>x</em>-intercept).

Such a triangle has area

1/2 × 12 × 12 = 72

and the area you want to cut away from this is given by a single integral,

\displaystyle \int_0^3 ((12-x)-x^2)\,\mathrm dx = \int_0^3(12-x-x^2)\,\mathrm dx

The integral has a value of

\displaystyle \left(12x-\frac{x^2}2-\frac{x^3}3\right)\bigg|_0^3 = 12\times3 - \frac{3^2}2 - \frac{3^3}3 \\\\ = \frac{45}2

and so the area of the shaded region is again 72 - 45/2 = 99/2.

7 0
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Help <br> Question in photo
Marysya12 [62]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
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