Answer:
Son's age is 25.
Step-by-step explanation:
Let the Father's age be 2x.
Let the son's age be x.
Five years ago:
Man's age will be = 2x - 5
Son's age will be = x - 5
If the ratio of their ages five years ago was 9:4, then:
2x - 5/x-5 = 9/4
⇒ 4(2x - 5) = 9(x - 5)
⇒ 8x - 20 = 9x - 45
⇒ -20 + 45 = 9x - 8x
⇒ 25 = x
∴ x = 25
Hence, the son's age = 25 years
If x = 25, the father's age (2x) will be 2 × 25 = 50 years
Answer:
The area of the square is 85 units^2
Step-by-step explanation:
Okay, here in this question, we are interested in calculating the area of the unknown square.
Kindly note that, since each of the other shapes are squares too, it means that the length of their sides is simply the square root of their areas.
Thus, the length of the squares are ;
√35 units and √50 units respectively
Now to find the area of the larger square, we employ the use of Pythagoras’ theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides
Let’s call the unknown length X
x^2 = (√35)^2 + (√50)^2
x^2 = 35 + 50
x^2 = 85
x = √85 units
Now as we know that the area of a square is simply the length of the side squared,
The area of the biggest square is simply (√85)^2 = 85 units^2
Answer:
(A)
Step-by-step explanation:
Cost of car = $21,349
Down payment = $3000
Remaining amount left = $18,349
Monthly payments (A) = $352
n = Total periods = 5years = 5*12= 60
P(loan amount) = 18,349
r = rate of interest monthly = r/12
Using formula, A = 
352 = [tex]\frac{18349\frac{r}{12} }{1-(1+\frac{r}{12})^{-60} }[/tex
= 0.059(approx)
Annual percentage rate = 5.9%
the idea behind the recurring decimal as a fraction, is to first off, multiply or divide by some power of 10, in order that we leave the recurring decimal to the right of the decimal point.
then we multiply by a power of 10, in order to move the repeating digits to the left of the decimal point, anyhow, let's proceed.
![\bf 0.6\overline{1212}\implies \cfrac{06.\overline{1212}}{10}\implies \cfrac{6+0.\overline{1212}}{10}\qquad \qquad \stackrel{\textit{now let's make}}{x=0.\overline{1212}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llll} 100\cdot x &=& 12.\overline{1212}\\\\ &&12+0.\overline{1212}\\\\ &&12+x\\\\ 100x&=&12+x\\\\ 99x&=&12\\\\ x&=&\cfrac{12}{99}\implies x = \cfrac{4}{33} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%200.6%5Coverline%7B1212%7D%5Cimplies%20%5Ccfrac%7B06.%5Coverline%7B1212%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2B0.%5Coverline%7B1212%7D%7D%7B10%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bnow%20let%27s%20make%7D%7D%7Bx%3D0.%5Coverline%7B1212%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20100%5Ccdot%20x%20%26%3D%26%2012.%5Coverline%7B1212%7D%5C%5C%5C%5C%20%26%2612%2B0.%5Coverline%7B1212%7D%5C%5C%5C%5C%20%26%2612%2Bx%5C%5C%5C%5C%20100x%26%3D%2612%2Bx%5C%5C%5C%5C%2099x%26%3D%2612%5C%5C%5C%5C%20x%26%3D%26%5Ccfrac%7B12%7D%7B99%7D%5Cimplies%20x%20%3D%20%5Ccfrac%7B4%7D%7B33%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \cfrac{06.\overline{1212}}{10}\implies \cfrac{6+x}{10}\implies \cfrac{6+\frac{4}{33}}{10}\implies \cfrac{~~\frac{202}{33}~~}{10}\implies \cfrac{~~\frac{202}{33}~~}{\frac{10}{1}}\implies \cfrac{202}{330} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{101}{165}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B06.%5Coverline%7B1212%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2Bx%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2B%5Cfrac%7B4%7D%7B33%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B202%7D%7B33%7D~~%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B202%7D%7B33%7D~~%7D%7B%5Cfrac%7B10%7D%7B1%7D%7D%5Cimplies%20%5Ccfrac%7B202%7D%7B330%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Ccfrac%7B101%7D%7B165%7D~%5Chfill)
notice, we first divided by 10, to move the decimal point over to the right by 1 slot, then we multiplied by 100, to move it two digits over the decimal point, namely the repeating "12", thus we use 100.
