<h3>The lateral area for the pyramid with the equilateral base is 144 square units</h3>
<em><u>Solution:</u></em>
The given pyramid has 3 lateral triangular side
The figure is attached below
Base of triangle = 12 unit
<em><u>Find the perpendicular</u></em>
By Pythagoras theorem
![hypotenuse^2 = opposite^2 + adjacent^2](https://tex.z-dn.net/?f=hypotenuse%5E2%20%3D%20opposite%5E2%20%2B%20adjacent%5E2)
Therefore,
![opposite^2 = 10^2 - 6^2\\\\opposite^2 = 100 - 36\\\\opposite^2 = 64\\\\opposite = 8](https://tex.z-dn.net/?f=opposite%5E2%20%3D%2010%5E2%20-%206%5E2%5C%5C%5C%5Copposite%5E2%20%3D%20100%20-%2036%5C%5C%5C%5Copposite%5E2%20%3D%2064%5C%5C%5C%5Copposite%20%3D%208)
<em><u>Find the lateral surface area of 1 triangle</u></em>
![\text{ Area of 1 lateral triangle } = \frac{1}{2} \times opposite \times base](https://tex.z-dn.net/?f=%5Ctext%7B%20Area%20of%201%20lateral%20triangle%20%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20opposite%20%5Ctimes%20base)
![\text{ Area of 1 lateral triangle } = \frac{1}{2} \times 8 \times 12\\\\\text{ Area of 1 lateral triangle } = 48](https://tex.z-dn.net/?f=%5Ctext%7B%20Area%20of%201%20lateral%20triangle%20%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%208%20%5Ctimes%2012%5C%5C%5C%5C%5Ctext%7B%20Area%20of%201%20lateral%20triangle%20%7D%20%3D%2048)
<em><u>Thus, lateral surface area of 3 triangle is:</u></em>
3 x 48 = 144
Thus lateral area for the pyramid with the equilateral base is 144 square units
Answer:
r = 26
Step-by-step explanation:
If you can graph a function, you can graph piecewise functions. Each one of them is a different line or curve within the domain for that specific line of curve. If the domain states it's less than or greater than a number, you circle that point on the line. If the domain states it's also possibly equal to the point at the beginning or the end, you make a closed dot.
I"m sure that's clear as mud, please respond with any additional questions.
Answer:
The diagram of the plotting point
is attached below.
Step-by-step explanation:
Given the points
![\left(3\frac{1}{2},\:2\frac{3}{4}\right)](https://tex.z-dn.net/?f=%5Cleft%283%5Cfrac%7B1%7D%7B2%7D%2C%5C%3A2%5Cfrac%7B3%7D%7B4%7D%5Cright%29)
as
![3\frac{1}{2}=\frac{7}{2}=3.5](https://tex.z-dn.net/?f=3%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7B7%7D%7B2%7D%3D3.5)
![2\frac{3}{4}=\frac{11}{4}=2.75](https://tex.z-dn.net/?f=2%5Cfrac%7B3%7D%7B4%7D%3D%5Cfrac%7B11%7D%7B4%7D%3D2.75)
so the point can be visualized as:
![\left(3\frac{1}{2},\:2\frac{3}{4}\right)=\left(3.5,\:2.75\right)](https://tex.z-dn.net/?f=%5Cleft%283%5Cfrac%7B1%7D%7B2%7D%2C%5C%3A2%5Cfrac%7B3%7D%7B4%7D%5Cright%29%3D%5Cleft%283.5%2C%5C%3A2.75%5Cright%29)
Now, we can check the point x = 3.5, and determine the corresponding value y = 2.75 and plot the point at the location (x, y) = (3.5, 2.75)
The diagram of the plotting point
is attached below.
Answer:
Given the function f(x) = 3x + 1, evaluation of f(a + 1) gives:
C. 3a + 4
Step-by-step explanation:
Given function:
f(x) = 3x + 1
We have to find f(a+1).
For this purpose, we will take x = a+1 and
substitute it in the function f(x) = 3x+1:
f(x) = 3x + 1
f(a+1) = 3(a+1) +1
f(a+1) = 3(a) + 3(1) +1
f(a+1) = 3a+3+1
f(a+1) = 3a + 4
So the function f(a+1) is equal to option C. 3a + 4.