<h3>Explanation:</h3>
<em>Lateral Area</em>
The lateral area is the area of the sides of the prism. If the faces are perpendicular to the bases, then each face is a rectangle. The area of each rectangle is the product of its length and width, generally the product of the height of the prism and the length of one edge of the base.
The total lateral area will then be the product of the height of the prism and the perimeter of the base.
<em>Total Area</em>
The total area is the sum of the lateral area (computed as above) and the area of the two bases of the prism. The formula for that area depends on the shape of the prism. (You have already seen formulas for the areas of triangles, rectangles, and other plane shapes. If not, they are readily available in your text or using a web search.)
Answer:
Step-by-step explanation:
hello :
−3 − (−2)= -3+2 = -1
Answer:
So do you need a solution?
Step-by-step explanation:
Hey there!
Three points to keep in mind!
- The given options are in standard form: ax+by=c
- The line is a solid line, hence, the possible outcomes would be ≤ or ≥
- The shaded part is above the line, hence, the symbol would be ≥
First, let's find the slope:
Slope : (y2-y1)/(x2-x1)
Slope : (0-3)/(2-0)
Slope : -3/2
Looking at line, we know that the y-intercept is 3 b=3
Slope intercept form:
<em>y≥(-3/2)x+3 </em>
Let's now convert this into a standard form:
<em>(3/2)x+y≥3 </em>
<em>Multiplying all the sides with 2: </em>
<em>Remember that the coefficient of 'x' shall not be negative or a fraction!</em>
<em>3x+2y≥6 </em><em>is the final answer!</em>
Answer:
![\left[\begin{array}{cc}-\frac{1}{2}&\frac{1}{6}\\-\frac{1}{2}&\frac{1}{3}\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B6%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B3%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If a 2x2 matrix is given as:
![\left[\begin{array}{cc}a&b\\c&d\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The inverse is:
![\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bad-bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Our matrix given is:
![\left[\begin{array}{cc}-4&2\\-6&6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-4%262%5C%5C-6%266%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<em />
<em>Using the formula, let's find the inverse:</em>
<em>![\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\\\end{array}\right]\\=\frac{1}{(-4)(6)-(-6)(2)}\left[\begin{array}{cc}6&-2\\6&-4\\\end{array}\right]\\=\frac{1}{-12}\left[\begin{array}{cc}6&-2\\6&-4\\\end{array}\right]\\=\left[\begin{array}{cc}-\frac{1}{2}&\frac{1}{6}\\-\frac{1}{2}&\frac{1}{3}\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bad-bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cfrac%7B1%7D%7B%28-4%29%286%29-%28-6%29%282%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-2%5C%5C6%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cfrac%7B1%7D%7B-12%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-2%5C%5C6%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B6%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B3%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
</em>
