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nadezda [96]
2 years ago
15

The equation C(l) = l3 – l2 + l + 2.5 models the cost to mail a package as a function of its length. To the nearest quarter inch

, approximately what is the length of a package that costs $11.00 to mail? Enter your answer as a decimal.
inches

Mathematics
2 answers:
Shalnov [3]2 years ago
7 0

Answer: 2.25

Step-by-step explanation:

Anettt [7]2 years ago
6 0

Answer:

Step-by-step explanation:

Set the cost equation C(l) = l3 – l2 + l + 2.5 equal to $11.00 and solve for l:

                                     C(l) = 2l + 2.5 = $11, or

                                               2l = 8.5, or l = length = 4.25 inches

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What is the y-int of the line given by the equation y=3x-11
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4 0
2 years ago
Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

3 0
3 years ago
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nirvana33 [79]

Answer:

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Step-by-step explanation:

Step 1:

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Step 1:

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9 × 6 = 54

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3 years ago
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