Answer: 1.8
<u>Step-by-step explanation:</u>
![\dfrac{2}{3}x-\dfrac{1}{5}>1\\\\\dfrac{2}{3}x\quad \quad >1+\dfrac{1}{5}\\\\\\\dfrac{2}{3}x\quad \quad >\dfrac{6}{5}\\\\\\\dfrac{2}{3}x\bigg(\dfrac{3}{2}\bigg) \ >\dfrac{6}{5}\bigg(\dfrac{3}{2}\bigg) \\\\\\x\qquad \quad >\dfrac{18}{10}\\\\\\x\qquad \quad >1.8](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B3%7Dx-%5Cdfrac%7B1%7D%7B5%7D%3E1%5C%5C%5C%5C%5Cdfrac%7B2%7D%7B3%7Dx%5Cquad%20%5Cquad%20%3E1%2B%5Cdfrac%7B1%7D%7B5%7D%5C%5C%5C%5C%5C%5C%5Cdfrac%7B2%7D%7B3%7Dx%5Cquad%20%5Cquad%20%3E%5Cdfrac%7B6%7D%7B5%7D%5C%5C%5C%5C%5C%5C%5Cdfrac%7B2%7D%7B3%7Dx%5Cbigg%28%5Cdfrac%7B3%7D%7B2%7D%5Cbigg%29%20%5C%20%3E%5Cdfrac%7B6%7D%7B5%7D%5Cbigg%28%5Cdfrac%7B3%7D%7B2%7D%5Cbigg%29%20%5C%5C%5C%5C%5C%5Cx%5Cqquad%20%5Cquad%20%3E%5Cdfrac%7B18%7D%7B10%7D%5C%5C%5C%5C%5C%5Cx%5Cqquad%20%5Cquad%20%3E1.8)
Answer:
Balance = $14,723.06
Step-by-step explanation:
Given:
Principal (P) = $15,000
Interest Rate (R) = 7.5%
Time (T) = ¹/12
Monthly payment = $185.49
Required:
Balance of the loan at the end of 3 months using the formula for simple interest.
SOLUTION:
![I = \frac{P.T.R}{100*12}](https://tex.z-dn.net/?f=%20I%20%3D%20%5Cfrac%7BP.T.R%7D%7B100%2A12%7D%20)
First Month:
Principal => $15,000
Add interest =>
=> $93.75
= $15,093.75
Less payment => $185.49
= $14,908.26
Second Month:
Principal => $14,908.26
Add interest =>
=> $93.18
= $15,001.44
Less payment => $185.49
= $14,815.95
Third Month:
Principal => $14,815.95
Add interest =>
=> $92.60
= $14,908.55
Less payment => $185.49
<h2>
Balance = $14,723.06</h2>
Answer:
It's b
Step-by-step explanation:
It is b or the second option because -2 is equal or greater than A therefore it is option 2
Answer:
< 1 = 88
< 2 = 92
Step-by-step explanation:
First add 4x - 4 + 4x = 180
X = 23
4(23) - 4 = 88
4(23) = 92
F(6) means the value of function at 6.
This means we have to check the value of function against the value which is 6. First column is y. The last row of the the table shows y = 6 and the value of function at this value of y.
So the value of function at y = 6 is f(6) = 3