Answer:
Step-by-step explanation:
<u>Use the law of cosines to find the side AB:</u>
<u>Use the Heron's area formula next:</u>
, where s- semi perimeter
- s = 1/2[x + x + 3 +
) = 1/2 (2x + 3 + ) - s - a = 1/2 (2x + 3 + - 2x - 6) = 1/2 (
- 3) - s - b = 1/2 (2x + 3 + - 2x) = 1/2 ( + 3)
- s - c = 1/2 (2x + 3 + - 2) = 1/2 (2x + 3 - )
<u>Now</u>
- (s - a)(s - b) = 1/4 [(x²+3x+9) - 9] = 1/4 (x² + 3x)
- s(s - c) = 1/4 [(2x + 3)² - (x² + 3x + 9)] = 1/4 (3x²+ 9x) = 3/4(x² + 3x)
<u>Next</u>
- A² = 3/16(x² + 3x)(x² + 3x)
- 300 = 3/16(x² + 3x)²
- 1600 = (x² + 3x)²
- x² + 3x = 40
<u>Substitute this into the first equation:</u>
Answer:
The fraction of blue shirts that she sold is 
Step-by-step explanation:
She sold:
1/3 on Tuesday.
5/12 on Wednesday.
What fraction of blue shirts did she sell on Tuesday and Wednesday?
Sum of fractions.
The lesser common multiplier between 3 and 12, which is 12.
So
The fraction of blue shirts that she sold is
$72. The tip is $0.2 per $1, so you would pay $12 in tips plus the original $60
Answer:
x= 42
y= 23
Step-by-step explanation:
Set up a system of equations like this:
(4y-9)=(3x-43)
(2x+13)=(4y+5)
when two lines cross like this they form vertical angles, and vertical angles are always equal.
calculate the system of equations and you'll get x=42 and y=23
To find the measure of any of the angles just put x or y into the equation.
Answer:
b. You would conclude that the differences in the average scores can be traced to differences in the working memory of the two groups.
Step-by-step explanation:
Though the average scores of the two sets could have lead to various conditions, but retentive ability deminishes with respect to an increase in age. With respect to the age of the elderly people involved, it is expected that some of them would not be able to retain information for a long period of time. Thus, their average score is 72%.
The college students' are younger, so it is expected that they should be able to retain more information. That ability is one of the reasons why their average score is 85%.
It can be concluded from the research that the differences in the average scores is probably due to the working memory of the two groups.
