Answer:
idk what you asking but if it solve for for x it going be no solution
I mean it would be 14.0 because it’s a whole number unless you are getting into 14 tenths or 14 hundredths then it’s like 0.14 or 0.014.
Answer:
![\sqrt[5]{13^3} = 13^{\frac{3}{5}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B13%5E3%7D%20%3D%2013%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
Step-by-step explanation:
Answers:
1) ![(3x+2)+(4-5x)=-2x+6](https://tex.z-dn.net/?f=%283x%2B2%29%2B%284-5x%29%3D-2x%2B6)
![(3+2i)+(4-5i)=7-3i](https://tex.z-dn.net/?f=%283%2B2i%29%2B%284-5i%29%3D7-3i)
2) ![(1-3x)(2+x)=-3x^{2}-5x+2](https://tex.z-dn.net/?f=%281-3x%29%282%2Bx%29%3D-3x%5E%7B2%7D-5x%2B2)
Step-by-step explanation:
In mathematics there are rules related to complex numbers, specifically in the case of addition and multiplication:
<u>Addition:
</u>
If we have two complex numbers written in their binomial form, the sum of both will be a complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imaginary parts (similarly as the sum of two binomials).
For example, the addition of these two binomials is:
![(3x+2)+(4-5x)=3x-5x+2+4=-2x+6](https://tex.z-dn.net/?f=%283x%2B2%29%2B%284-5x%29%3D3x-5x%2B2%2B4%3D-2x%2B6)
Similarly, the addition of two complex numbers is:
Here the complex part is the number with the ![i](https://tex.z-dn.net/?f=i)
<u>Multiplication:
</u>
If we have two complex numbers written in their binomial form, the multiplication of both will be the same as the multiplication (product) of two binomials, taking into account that
.
For example, the multiplication of these two binomials is:
![(1-3x)(2+x)=2+x-6x-3x^{2}=-3x^{2}-5x+2](https://tex.z-dn.net/?f=%281-3x%29%282%2Bx%29%3D2%2Bx-6x-3x%5E%7B2%7D%3D-3x%5E%7B2%7D-5x%2B2)
Similarly, the multiplication of two complex numbers is:
Answer: B
$10,390.75
Step-by-step explanation:
Given that the lease = $291 per month
Lease pay = 291 × 24 = $6984
Time = 24 months
down payment = $458.
The lease allows for 12,000 miles per year and includes a $0.35 per mile charge for miles driven in excess of that amount.
For the car to be driven a total of 32,425 miles.
Excess = 32425 - 24000 = 8425 miles
Charges = 0.35 × 8425 = $2948.75
The cost for two years for this vehicle will be
2948.75 + 458 + 6984 = $10390.75