If 3+4 = 7 and 6+7 = 7 leads to 3+4 = 6+1, then you are using the transitive property. In general the property says that if a = b and b = c, then a = c.
Answer:
✑ 
Let us consider , there are 30 boys and 40 girls in a school. So, the ratio of number of boys to number of girls =
and read as 3 is to 4. We can say , the ratio compares two of more quantities of same unit. Until two quantities have same units , it is meaningless to compare by ratio. So, to find the ratio of two quantities , it is necessary to express them in same unit or of same kind. Since , they are in division form. So ,ratio is unit less quantity. Thus , the ratio is a comparison of two quantities of the same kind in division which doesn't have any units. For example :
is a ratio and is read as ' a is to b ' in which ' a : is antecedent and ' b ' is called consequent.
✎ 
☪ 
£ 21 is to be divided between Amy and Ben in the ratio 5 : 2. So, let Amy get £ 5x and Ben get £ 2x.
Then , According to the question , 
Solve for x :
⇝ 
⇝ 
⇝ 
The value of x is 3. Now , substitute the value of x in 5x :
Amy received £ 5x = 5 × 3 = £ 15 .
☥ 
Hope I helped ! ツ
Have a wonderful day / night ♡
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8*10*14 = 1120
10*8*10 = 800
1120+800 = 1920
volume =1920 cm3
Question 7
(37,065 - 26, 102)/(37, 065) = P/100
(10, 963)/(37, 065) = P/100
37, 065P = (10, 963)(100)
37,065P = 109, 630
P = 109, 630 ÷ 37, 065
P = 2.9577768785
P = 295.78%
Do likewise for the rest of the questions.
To change from decimal to percent, move the decimal point two places to the right and then add the percent symbol (%).
Answer:
<h3><u>Option 1</u></h3>
Earn $50 every month.
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000

This is a <u>linear function</u>.
<h3><u>Option 2</u></h3>
Earn 3% interest each month.
(Assuming the interest earned each month is <u>compounding interest</u>.)
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000

This is an <u>exponential function</u>.
<h3><u>Table of values</u></h3>
<u />

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.
However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.
From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.
<h3><u>Conclusion</u></h3>
If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.