2. Dividing exponential expression with the same base is what the left hand fractional expression is. Z^8/Z^?=z^6. Since all the bases are the same. To get the exponential power 6 on the right we subtract the denominator’s (bottom part of the fraction) exponential power (on the left) from the numerator’s (top part) exponential power.
Then, 8 - ? = 6? 2 is the only number to make the left equal the right side. This, z^8/z^2=z^6
The right answer is Option D.
Step-by-step explanation:
Given equations are;
2x+4y=3 Eqn 1
x+3y=13 Eqn 2
When we use subtraction-addition method, we make one of the variables same with opposite signs so that only one variable remains after addition or subtraction.
In the given problem, we will multiply Eqn 2 with "-2" so that the x variables become equal and then we can add both the equations and solve for y.
Therefore,
The first step will be to multiply the second equation by -2 to solve the linear system of equations.
The right answer is Option D.
Keywords: linear equations, subtraction
Learn more about linear equations at:
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After 7 years fund balance will be $2152336.05 if the philanthropist pledges to donate 10% of a fund each year.
<h3>What is an exponential function?</h3>
It is defined as the function that rapidly increases and the value of the exponential function is always a positive. It denotes with exponent. 
where a is a constant and a>1
We have:
A philanthropist pledges to donate 10% of a fund each year.
We can find the amount left after 7 years as follows:
A = 450000(1 - 0.1)⁷
10% of a fund each year.
Rate r = 10% = 0.1
A = 450000(0.9)⁷
A = $2152336.05
Thus, after 7 years fund balance will be $2152336.05 if the philanthropist pledges to donate 10% of a fund each year.
Learn more about the exponential function here:
brainly.com/question/11487261
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Given a complex number in the form:
![z= \rho [\cos \theta + i \sin \theta]](https://tex.z-dn.net/?f=z%3D%20%5Crho%20%5B%5Ccos%20%5Ctheta%20%2B%20i%20%5Csin%20%5Ctheta%5D)
The nth-power of this number,

, can be calculated as follows:
- the modulus of

is equal to the nth-power of the modulus of z, while the angle of

is equal to n multiplied the angle of z, so:
![z^n = \rho^n [\cos n\theta + i \sin n\theta ]](https://tex.z-dn.net/?f=z%5En%20%3D%20%5Crho%5En%20%5B%5Ccos%20n%5Ctheta%20%2B%20i%20%5Csin%20n%5Ctheta%20%5D)
In our case, n=3, so

is equal to
![z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20%5Crho%5E3%20%5B%5Ccos%203%20%5Ctheta%20%2B%20i%20%5Csin%203%20%5Ctheta%20%5D%20%3D%20%285%5E3%29%20%5B%5Ccos%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%2B%20i%20%5Csin%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%5D)
(1)
And since

and both sine and cosine are periodic in

, (1) becomes