Answer:
1. In the multiplication of the imaginary parts, the student forgot to square of i. OR
2. The student has only multiplied the real parts and the imaginary parts.
Correct value
.
Step-by-step explanation:
The given expression is
![(4+5i)(3-2i)](https://tex.z-dn.net/?f=%284%2B5i%29%283-2i%29)
A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i.
Student's mistake can be either 1 or second:
1. In the multiplication of the imaginary parts, the student forgot to square of i.
2. The student has only multiplied the real parts and the imaginary parts.
![(4+5i)(3-2i)=4\times 3+5\times (-2)i=12-10i](https://tex.z-dn.net/?f=%284%2B5i%29%283-2i%29%3D4%5Ctimes%203%2B5%5Ctimes%20%28-2%29i%3D12-10i)
Which is not correct. The correct steps are shown below.
Using distributive property, we get
![4(3-2i)+5i(3-2i)](https://tex.z-dn.net/?f=4%283-2i%29%2B5i%283-2i%29)
![4(3)+4(-2i)+5i(3)+5i(-2i)](https://tex.z-dn.net/?f=4%283%29%2B4%28-2i%29%2B5i%283%29%2B5i%28-2i%29)
![12-8i+15i-10i^2](https://tex.z-dn.net/?f=12-8i%2B15i-10i%5E2)
![[\because i^2=-1]](https://tex.z-dn.net/?f=%5B%5Cbecause%20i%5E2%3D-1%5D)
![12+7i+10](https://tex.z-dn.net/?f=12%2B7i%2B10)
![22+7i](https://tex.z-dn.net/?f=22%2B7i)
Therefore, the correct value of
is
.
Using the surface area formula for rectangular and triangular prism, the surface area of the composite figure is: 444 m².
<h3>What is the Surface Area of the Composite Figure?</h3>
Total surface area = surface area of the top triangular prism + surface area of the bottom rectangular prism - area of the surface both share together.
Surface area of the top triangular prism = (S1 + S2+ S3)L + bh = (10 + 10 + 16)5 + (16)(6) = 276 m².
Surface area of the bottom rectangular prism = 2(wl + hl + hw) = 2·(5·16+4·16+4·5) = 328 m²
Area of the surface both share together = 2(16)(5) = 160 m²
Total surface area = 276 + 328 - 160 = 444 m².
Learn more about the surface area of triangular prism on:
brainly.com/question/16147227
#SPJ1
Answer:
2/3
Step-by-step explanation:
The slope of any function is its first derivative.
If f(x)=x^2, then f'(x)= 2x .
At x=5, slope = f'(x) = 2(5) = <em>10</em> .