Answer:
x= 62
y= 54
Step-by-step explanation:
Step one:
given data
let the numbers be x and y and the larger be x the smaller be y
The difference between two numbers is 8
x-y= 8-----------1
If the larger is subtracted from three times the smaller, the difference is 100
3y-x=100------------2
from eqn 1, x= 8+y
put this in eqn 2
3y-(8+y)=100
3y-8-y=100
collect liker terms
3y-y-8=100
2y=108
y= 54
put y= 54 in eqn 1
x-y=8
x-54= 8
x= 8+54
x=62
The surface area of the rectangular prism with the dimensions that are stated is: 384 in.²
<h3>What is the Surface Area of a Triangular Prism?</h3>
Surface area = perimeter of base × height of prism + 2(base area)
= (s1 + s2 + s3)L + 2(1/2bh)
Given the following:
- side of base (s1) = 6 in.
- side of base (s2) = 8 in.
- side of base (s3) = 10 in.
- Length of prism (L) = 14 in.
- Triangular base length (b) = 6 in.
- h = 8 in.
Surface area = (6 + 8 + 10)14 + 2(1/2 × 6 × 8)
Surface area = 384 in.²
Learn more about the surface area of a rectangular prism on:
brainly.com/question/1310421
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Answer:
![\large\boxed{12\sqrt5}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B12%5Csqrt5%7D)
Step-by-step explanation:
![\text{Use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\-----------------\\\\-4\sqrt{15}\cdot\left(-\sqrt3\right)=4\sqrt{(15)(3)}=4\sqrt{5\cdot3\cdot3}=4\sqrt5\cdot\sqrt9\\\\=4\sqrt5\cdot3=12\sqrt5](https://tex.z-dn.net/?f=%5Ctext%7BUse%7D%5C%20%5Csqrt%7Bab%7D%3D%5Csqrt%7Ba%7D%5Ccdot%5Csqrt%7Bb%7D%5C%5C-----------------%5C%5C%5C%5C-4%5Csqrt%7B15%7D%5Ccdot%5Cleft%28-%5Csqrt3%5Cright%29%3D4%5Csqrt%7B%2815%29%283%29%7D%3D4%5Csqrt%7B5%5Ccdot3%5Ccdot3%7D%3D4%5Csqrt5%5Ccdot%5Csqrt9%5C%5C%5C%5C%3D4%5Csqrt5%5Ccdot3%3D12%5Csqrt5)
Answer:
F (I think)
Step-by-step explanation:
2/3 is 0.66 (forever), 65%, 5/8 is .625, and 0.6 (0.600)
Hopefully, this helps :D
The mean absolute deviation is the average of the average. It shows the average distance of the numbers in a data set from the mean. After finding the mean, you find the absolute value of the distance between each number, and find the mean of those numbers to find the mean absolute deviation.