Answer:
a. We spent all together 40.2+24.5+37.8=$102.50.
b. We saved 19.8+10.5+7.2=$37.50
c. 
Step-by-step explanation:
To find the amount we spend, we multiply the original price by the decimal conversion of each fraction. Then we add the amounts together.
- Sneakers cost $60 and you bought them at 2/3=0.6666....
60(0.67)=$40.20 and saved $19.80
- Sweater costs $35 and you bought it at 7/10=0.7
35(0.7)=$24.50 and saved $10.50
- A jacket costs $45 and you bought it at 5/6=0.8333...
45(0.84)=$37.80 and saved $7.20
We spent all together 40.2+24.5+37.8=$102.50.
We saved 19.8+10.5+7.2=$37.50
We could of spent paying regular price 60+35+45=$140. To find the fraction we place the total we actually paid over the total regular price.
In 36 days will the science and math teachers both give tests on the same day again.
<u>Step-by-step explanation:</u>
Here we have , The 6th grade science teacher gives a test every 12 days and the math teacher gives a test every 9 days. Today, both the science and math are giving tests. We need to find In how many days will the science and math teachers both give tests on the same day again . Let's find out:
It's given that science teacher gives a test every 12 days and the math teacher gives a test every 9 days . In order to find the number of days after which they'll give test at same day , we will find LCM of 12 & 9 . i.e.
⇒ 
and ,
⇒ 
Hence , LCM of 12 and 9 is 36 . Therefore , In 36 days will the science and math teachers both give tests on the same day again.
Answer:
$1880.48 to the nearest cent.
Step-by-step explanation:
Total after n years = P(1 + r)^n where P = initial deposit, r = the rate as a fraction.
4 5/8 % = 0.04625.
Here the total after 5 years = 1500(1 + 0.04625)^5
= $1880.48 to the nearest cent.
Hello kiddo.
<span>From least to greatest 1/5, 0.1, -1/2, -0.25, 0.3
</span>
-1/2 , -0.25, 1/5, 0.3, 0.1
Answer:
E) x2 + 10x + 25
Step-by-step explanation:
A number is increased by 5, and then squared is
(x + 5)^2
When taken out of the parenthesis it is:
x2 + 10x + 25 is your answer.
