If he charges $145 he worked for 5 hours at $20 an hour
<h2>~<u>Solution</u> :-</h2>
Here, it is given that the bag contains 25 paise coins and 50 paise coins in which, 25 paise coins are 6 times than that of 50 paise coins. Also, the total money in the bag is Rs. 6.
- Hence, we can see that, here, we have been given the linear equation be;
Let the number of coins of 50 paise will be $ x $ and the number of coins of 25 paise will be $ 6x $ as given. . .
Hence,
- Hence, the number of 50 paise coins will be <u>2</u>. And, 6 times of two be;
- Hence, the number of 25 paise coins will be <u>12</u>.
Let m = slope = -4/5
P(x,y) = P(5,10)
Slope-intercept form for the equation of a line:
y=mx+c
Plug m= -4/5 into the slope-intercept formula
y = -4/5x + c
Plug P(5,10) into point-slope formula
y = -4/5x + c
10 = -4/5(5) + c
10 = -4 + c
substract -4 from both sides,
10 - (-4) = -4 - (-4) + c
14 = c
So, the equation is y = -4/5x + 14
Add two 0's to 14.7 making 14.700
Answer:
139,999
Step-by-step explanation:
If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999
