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Sholpan [36]
2 years ago
5

I need help, I don't quite understand

Mathematics
2 answers:
maksim [4K]2 years ago
5 0

Answer:

1

Step-by-step explanation:

so we have:

f+2>5 2 goes the other side, so

f>5-2

f>3

f+4<9 4 goes to the other side, so

f<9-4

f<5

we have two inequality

f>3

f<5

f belongs (3,5), we have only one value for this, it is 4

OverLord2011 [107]2 years ago
3 0
F can be equal to both 1 and 2. Reason why: If you input 1 or 2 to replace f on either inequality, it will never be greater than or equal to 5 or 9.
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Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
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<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.

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