Answer:
Alexander is incorrect because the expressions are not equivalent.
Step-by-step explanation:
If the expression is evaluated for any value of x, y; the result will not be same.
For instance, let assume x = 1 and y = 2
3x + 4y = 3 + 4 = 7
(3)(4) + xy = (3)(4) + (1 * 2) = 12 + 2 = 14
So, the expressions are not the same and Alexander is incorrect.
Answer:
a. a[1] = 3; a[n] = 2a[n-1]
b. a[n] = 3·2^(n-1)
c. a[15] = 49,152
Step-by-step explanation:
Each term of the given sequence is 2 times the previous term. (This description is the basis of the recursive formula.) That is, the terms of the given sequence have a common ratio of 2. This means the sequence is geometric, so the formulas for explicit and recursive rules for a geometric sequence apply.
The first term is 3, and the common ratio is 2.
<h3>(a)</h3>
The recursive rule is ...
a[1] = 3
a[n] = 2×a[n-1]
__
<h3>(b)</h3>
The explicit rule is ...
a[n] = a[1]×r^(n-1)
a[n] = 3×2^(n-1)
__
<h3>(c)</h3>
The 15th term is ...
a[15] = 3×2^(15-1) = 3×2^14
a[15] = 49,152
<h3>
Answer: Choice A</h3>
![x^2\left(\sqrt[4]{x^2}\right)](https://tex.z-dn.net/?f=x%5E2%5Cleft%28%5Csqrt%5B4%5D%7Bx%5E2%7D%5Cright%29)
=====================================================
Explanation:
The fourth root of x is the same as x^(1/4)
I.e,
![\sqrt[4]{x} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%20%3D%20x%5E%7B1%2F4%7D)
The same applies to x^10 as well
![\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%5Cleft%28x%5E%7B10%7D%5Cright%29%5E%7B1%2F4%7D)
Multiply the exponents 10 and 1/4 to get 10/4
![\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%5Cleft%28x%5E%7B10%7D%5Cright%29%5E%7B1%2F4%7D%20%3D%20x%5E%7B10%2A1%2F4%7D%20%3D%20x%5E%7B10%2F4%7D)
![\sqrt[4]{x^{10}} = x^{10/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20x%5E%7B10%2F4%7D)
-----------------------
If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below
![x^{m/n} = x^a\sqrt[n]{x^b}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20x%5Ea%5Csqrt%5Bn%5D%7Bx%5Eb%7D)
The 'a' and 'b' are found through dividing m/n
m/n = a remainder b
'a' is the quotient, b is the remainder
-----------------------
The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4
m/n = 10/4 = 2 remainder 2
We have a = 2 and b = 2
So
turns into
![x^{10/4} = x^2\sqrt[4]{x^2}](https://tex.z-dn.net/?f=x%5E%7B10%2F4%7D%20%3D%20x%5E2%5Csqrt%5B4%5D%7Bx%5E2%7D)
which means
![\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%7Bx%5E2%7D%20%5Csqrt%5B4%5D%7Bx%5E2%7D)
The base is a square, so the area of the square is (10 cm)² = 100 cm²
To find the area of each triangle use the formula
base • height
10 cm • 9 cm = 90 cm²
now multiply by the number of triangles
90 cm • 4 = 360 cm²
Add the area of the square:
100 cm² + 360 cm² = 460 cm²
I hope it's correct
85 is composite because it can be divided by 5 and 17
