Answer:
Keith's unit rate of change of dollars with respect to time is $1500.
Step-by-step explanation:
It is given that Keith is saving money for a car.
Year 1: 1500
year 2: 3000
year 3: 4500
Let y be the saved amount after x year.
The coordinate pairs according to the given table are (1,1500), (2,3000) and (3,4500).
The formula for rate of change is
Consider any two coordinate pairs.
Let as consider (1,1500) and (2,3000). So, Keith's unit rate of change of dollars with respect to time is
Therefore, Keith's unit rate of change of dollars with respect to time is $1500.
Area = length of base × altitude
34 = 8.5 × a
34 = 8.5a
a = 34/8.5
a = 4
The altitude of the corresponding side is 4 cm
Problem 1
<h3>Answer: 7.3</h3>
Explanation: Apply the square root to the area to get the side length. This only applies to areas that are squares (hence the name).
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Problem 2
<h3>Answer: C) 1.3</h3>
Explanation: Use your calculator to find that choices A,B,D plugged into the square root function yield terminating decimal values. "Terminating" means "stop". This implies that they are perfect squares (though not perfect squares in the sense of whole number perfect squares which you may be used to). Choice C is the only value that has a square root that leads to a non-terminating decimal. The digits of this decimal go on forever without any pattern. The value is irrational.
- sqrt(5.29) = 2.3 terminating decimal
- sqrt(13.69) = 3.7 terminating decimal
- sqrt(1.3) = 1.140175425 keeps going forever without any pattern
- sqrt(0.09) = 0.3 terminating decimal
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Problem 3
<h3>Answer: 23.6 feet approximately</h3>
Explanation: Apply the square root to 15.5 to get roughly 3.937; this is the approximate side length of one square. Six of these tiles placed together will lead to a total length of roughly 6*3.937 = 23.622 which rounds to 23.6 feet. Like with problem 1, the square root being used like this only works for square areas.
14 11/18
I got this answer by doing 30 1/9 + 4 1/3= 34 4/9
Now I subtracted 34 4/9 by 19 5/6 to get the answer 14 11/18
TIP: When adding/subtracting/multiplying/dividing mixed numbers change them to an improper fraction first by multiplying the denominator and the whole number then adding the numerator
Check the picture below.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-9%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%20~%5Chfill%20AB%3D%5Csqrt%7B%5B%208-%201%5D%5E2%20%2B%20%5B%200-%20%28-9%29%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AB%3D%5Csqrt%7B7%5E2%2B%280%2B9%29%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B7%5E2%2B9%5E2%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B130%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}](https://tex.z-dn.net/?f=B%28%5Cstackrel%7Bx_1%7D%7B8%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B9%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%20~%5Chfill%20BC%3D%5Csqrt%7B%5B%209-%208%5D%5E2%20%2B%20%5B%20-8-%200%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20BC%3D%5Csqrt%7B1%5E2%2B%28-8%29%5E2%7D%5Cimplies%20%5Cboxed%7BBC%3D%5Csqrt%7B65%7D%7D)
now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.
