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bagirrra123 [75]
4 years ago
13

The Lamaster family is a family of five people. They have used 4,885.78 gallons of water so far this month. They cannot exceed 7

,250.50 gallons per month during drought season. Write an inequality to show how much water just one member of the family can use for the remainder of the month, assuming each family member uses the same amount of water every month.
5x − 4,885.78 ≤ 7,250.50
5x + 4,885.78 ≤ 7,250.50
5x − 4,885.78 ≥ 7,250.50
5x + 4,885.78 ≥ 7,250.50
Mathematics
1 answer:
oksian1 [2.3K]4 years ago
8 0

Let X be the amount of water one person can use.

Multiply that by 5 people to get 5X

That amount plus what they already used ( 4885.78) needs to be less than or equal to the amount they are limited to ( 7250.50 ).

Now write out the equation to get : 5x + 4885.78 ≤ 7250.50

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3 years ago
Consider the radical equation √c+22 = c + 2.
dimulka [17.4K]

First square both sides to get:

c + 22 = (c + 2)^2

or

c + 22 = c^2 + 4c + 4

Move the terms on the left side to the right side:

c^2 + 3c - 18 = 0

Factor to get:

(c + 6) * (c - 3) = 0.

The solutions are c = -6 and c = 3.

Check to see if these answers work by plugging them into the original equation:

c = -6:

sqrt (-6 + 22) ?= -6 + 2

But, -6 + 2 is a negative number, and you can't get a negative from a square root. So, -6 is extraneous.

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7 0
3 years ago
A bank offers all savings accounts 5% interest compounded annually. If one account has a principal of $100 and another
arsen [322]

Answer:

From the calculation for both the accounts, it is clear that both the account double in the same time period of 14 years 26 days .

Step-by-step explanation:

Given as :

The principal for the first account = p = $100

The rate of interest = r = 5% compounded annually

The account gets double , so, Amount = A = $200

Let the time after which account gets double = t years

So<u>, From Compound Interest method</u>

Amount = Principal × (1+\dfrac{\trxtrm rate}{100})^{\textrm time}

As amount is double its principal

So, A = 2 × $100 = $200

Or, A = p × (1+\dfrac{\trxtrm r}{100})^{\textrm t}

Or, $200 = $100 × (1+\dfrac{\trxtrm 5}{100})^{\textrm t}

Or, \dfrac{200}{100} = (1.05)^{\textrm t}

Or, 2 = (1.05)^{\textrm t}

Taking Log both side

Log_{10}2 = Log_{10}(1.05)^{t}

Or, 0.3010 = t Log_{10}1.05

Or, 0.3010 = t × 0.0211

∴ t = \dfrac{0.3010}{0.0211}

I.e t = 14.26

So, The time period to get account double is 14 years 26 days

<u>Again </u>

Amount = Principal × (1+\dfrac{\trxtrm rate}{100})^{\textrm time}

Or, A = p × (1+\dfrac{\trxtrm r}{100})^{\textrm t}

As amount is double its principal

So, A = 2 × $1000 = $2000

Or, $2000 = $1000 × (1+\dfrac{\trxtrm 5}{100})^{\textrm t}

Or, \dfrac{2000}{1000} = (1.05)^{\textrm t}

Or, 2 = (1.05)^{\textrm t}

Taking Log both side

Log_{10}2 = Log_{10}(1.05)^{t}

Or, 0.3010 = t Log_{10}1.05

Or, 0.3010 = t × 0.0211

∴ t = \dfrac{0.3010}{0.0211}

I.e t = 14.26

So, The time period to get account double is 14 years 26 days

Hence From the calculation for both the accounts, it is clear that both the account double in the same time period of 14 years 26 days . Answer

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Answer:

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Step-by-step explanation:

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3 years ago
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