Answer:
8y = 7x
Step-by-step explanation:
9. ¹/₃(x + 6) = 8
¹/₃(x) + ¹/₃(6) = 8
¹/₃x + 2 = 8
<u> - 2 - 2</u>
3 · ¹/₃x = 6 · 3
x = 18
15. ¹/₅(x + 10) = 6
¹/₅(x) + ¹/₅(10) = 6
¹/₅x + 2 = 6
<u> - 2 - 2</u>
5 · ¹/₅x = 4 · 5
x = 20
20. ¹/₈(24x + 32) = 10
¹/₈(24x) + ¹/₈(32) = 10
3x + 4 = 10
<u> - 4 - 4</u>
<u>3x</u> = <u>6</u>
3 3
x = 2
32. 5 - ¹/₂(x - 6) = 4
5 - ¹/₂(x) - ¹/₂(-6) = 4
5 - ¹/₂x + 3 = 4
5 + 3 - ¹/₂x = 4
8 - ¹/₂x = 4
<u>- 8 - 8</u>
-2 · (-¹/₂x) = -4 · (-2)
x = 8
33. ²/₃(3x - 6) = 3
²/₃(3x) - ²/₃(6) = 3
2x - 4 = 3
<u> + 4 + 4</u>
<u>2x</u> = <u>7</u>
2 2
x = 3¹/₂
Answer:
£160,000
Step-by-step explanation:
Alice bought the house in 2008 for £x.
In 2014, the house sold for a 20% profit, so it sold for 1.2x. (* see explanation below)
In 2019, the house sold for a 5% loss, so it sold for 0.95(1.2x). (* see explanation below)
In 2019, it sold for £182,400.
0.95(1.2x) = 182,400
1.14x = 182,400
x = 182,400/1.14
x = 160,000
Answer: £160,000
* Explanation of 1.2 and 0.95.
A quick way to apply a percent increase or decrease to a number is to using the following method. The number you start with is 100% of the number. If you are adding a percent to it, then add the percent to 100% and convert it to a decimal. If you are subtracting the percent, then subtract the percent from 100% and convert to a decimal. Then multiply the decimal by the original number to find the increased or decreased number.
Example of percent increase:
A book used to cost £10. The price went up by 12%. What is the new price?
100% + 12% = 112% = 1.12
1.12 * £10 = £11.20
Example of percent decrease:
A pair of trousers costs £60. It is on sale at 35% off the regular price. What is the sale price?
100% - 35% = 65% = 0.65
0.65 * £60 = £39
The answer is 21/32 that is the answer of 3/4 ×7/8
I think you forgot to give the options along with the question. I am answering the question based on my research and knowledge. "Some customers who rate their service as high quality return for additional services" is the conclusion that can be drawn from this study. I hope that the answer has actually come to your desired help.
