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Alika [10]
3 years ago
11

Leslie just switched to a new phone plan. Her old phone plan was $10 plus 10 cents per gigabyte of data overage. Her new plan is

$12 plus 8 cents per gigabyte of data overage. Leslie claims that she is going to pay less on her new plan because she always goes over her data limit. Her data overage was 14 gigabytes. Is Leslie's claim true?
Mathematics
1 answer:
Lilit [14]3 years ago
8 0
Y=10x + 10
Y= 8x + 12

Now just plug in 14 for x into both equations.
Y= 10(14) + 10 :150
Y= 8(14) + 12 :124
Yes Leslie’s claim is true
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Answer:

do you need help?

Step-by-step explanation:

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List the next four multiples of the fraction 7/12
choli [55]
<h2>Hello!</h2>

The answer is:

\frac{14}{12},\frac{21}{12},\frac{28}{12},\frac{35}{12}

<h2>Why?</h2>

Let's explain with an example the definition of a multiple.

Multiple of the number 1 are: 1, 2, 3, 4, 5 and so...

Multiple of the number 3 are: 3, 6, 9, 12, 15 and so...

A multiple of a number is the repeated sum of itself, from the example:

Multiple of the number 1 are: Itself, (1+1), (1+1+1), (1+1+1+1), (1+1+1+1+1) and so...

Therefore,

Multiple of \frac{7}{12} are:

\frac{7}{12} =\frac{7}{12}(itself)=\frac{7}{12}*1 \\\\\frac{14}{12}= \frac{7}{12}+\frac{7}{12}=\frac{7}{12}*2\\\\\frac{21}{12}=\frac{7}{12}+\frac{7}{12}+\frac{7}{12}=\frac{7}{12}*3\\\\\\\frac{28}{12}=\frac{7}{12}+\frac{7}{12}+\frac{7}{12}+\frac{7}{12}=\frac{7}{12}*4\\\\\frac{35}{12}=\frac{7}{12}+\frac{7}{12}+\frac{7}{12}+\frac{7}{12}+\frac{7}{12}=\frac{7}{12}*5

Have a nice day!

4 0
3 years ago
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Answer:

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according to the given figure.

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and x Is 5

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8 0
3 years ago
Read 2 more answers
An open box is to be constructed so that the length of the base is 3 times larger than the width of the base. If the cost to con
natta225 [31]

Answer:

Step-by-step explanation:

Let assume that the volume = 88 cubic feet

Then:

L = 3w --- (1)volume = l \times w \times h \\ \\ 88 =  (3w)   \times w  \times h \\ \\ 3w^2 h= 88 \\ \\ h = \dfrac{88}{3w^2}--- (2) \\ \\

The construction cost now is:

C = 4(l \times w) + 2 ( w  \times h) + 2(l \times h) \\ \\ C = 4(3w^2) + 2(w  \times \dfrac{88}{3w^2}) + 2(3w \times \dfrac{88}{3w^2}) \\\\ C = 12w^2 + \dfrac{176}{3w} + \dfrac{176}{w}

Now, to determine the minimum cost:

\dfrac{dC}{dw}= 0  \\ \\ \implies 24 w - \dfrac{176}{3w^2}- \dfrac{176}{w^2}=0 \\ \\ 24 w ^3 = \dfrac{176}{w^2}(\dfrac{1}{3}+1) \\ \\ 24 w ^3 = \dfrac{176(4)}{3} \\ \\ w^3 = \dfrac{88}{3(3)}

w = \dfrac{88}{3(3)}^{1/3} \ feet

Now;

length = 3 ( \dfrac{88}{3(3)})^{1/3} \ feet

height = ( 88})^{1/3} (3)^{1/3} \ feet

6 0
3 years ago
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