A. 60(2/3) + 35(7/10) + 45(5/6) = $102.00
b. 60 + 35 + 45 = 140 = original price
140 - 102 = $30 saved
c. 102/140 = .7286 * 100 = 72.86 % (not sure how you need it rounded)
Answer:
1/8
Step-by-step explanation:
To simplify the expression √3/√8, we can first simplify the square root terms by finding the prime factorization of each number under the square root. The prime factorization of 3 is 3, and the prime factorization of 8 is 2 * 2 * 2.
We can then rewrite the square root terms as follows:
√3/√8 = √(3) / √(2 * 2 * 2)
Next, we can use the property of square roots that says that the square root of a number is equal to the square root of each of its prime factors. This means that we can rewrite the square root term as follows:
√(3) / √(2 * 2 * 2) = √(3) / √(2) / √(2) / √(2)
Since the square root of a number is the same as the number itself, we can simplify the expression further by removing the square root symbols from the prime numbers 2:
√(3) / √(2) / √(2) / √(2) = √(3) / 2 / 2 / 2
Finally, we can use the rules of division to simplify the expression even further:
√(3) / 2 / 2 / 2 = √(3) / (2 * 2 * 2)
Since any number divided by itself is equal to 1, we can simplify the expression one last time to get our final answer:
√(3) / (2 * 2 * 2) = 1/2 * 1/2 * 1/2 = 1/8
Therefore, the simplified form of the expression √3/√8 is 1/8.
Answer:
1/5x+51=2x+42
Step-by-step explanation:
1/5 of x, the number, plus 51 is equal to 2 times the number plus 42.
<span>
<span><span>
x
y=2*(0.5)^x
</span><span>-10
2048
</span>
<span>
-9
1024
</span>
<span>
-8
512
</span>
<span>
-7
256
</span>
<span>
-6
128
</span>
<span>
-5
64
</span>
<span>
-4
32
</span>
<span>
-3
16
</span><span>-2
8
</span>
<span>
-1
4
</span>
<span>
0
2
</span>
<span>
1
1
</span>
<span>
2
0.5
</span>
<span>
3
0.25
</span>
<span>
4
0.125
</span>
<span>
5
0.0625
</span>
<span>
6
0.03125
</span>
<span>
7
0.015625
</span>
<span>
8
0.0078125
</span>
<span>
9
0.00390625
</span>
<span>
10
0.00195313
As x goes to negative infinity the function grows to infinity.
As x grows to infinity the function decreases an approximate to zero. </span></span></span>
