Answer:
<u>1728 in²</u>
Step-by-step explanation:
For the area of B, you need to find the length and width (height and base, whatever you want to call it).
For the height, you would need to do: 144 - 36 - 24 - 36 = <u>48 inches</u>
For the base, you would need to do: 60 - 24 = <u>36 inches</u>
Now you multiply the base and height to get the area: 48 x 36 = <u>1728 in²</u>
Answer:
1
Step-by-step explanation:
One. For example, if we chose 5 as the number, then the multiplicative inverse would be 1/5. Multiplying 5 by 1/5 results in one (1).
If you know how to solve word problems involving the sum of consecutive even integers, you should be able to easily solve word problems that involve the sum of consecutive odd integers. The key is to have a good grasp of what odd integers are and how consecutive odd integers can be represented.
Odd Integers
If you recall, an even integer is always 22 times a number. Thus, the general form of an even number is n=2kn=2k, where kk is an integer.
So what does it mean when we say that an integer is odd? Well, it means that it’s one less or one more than an even number. In other words, odd integers are one unit less or one unit more of an even number.
Therefore, the general form of an odd integer can be expressed as nn is n=2k-1n=2k−1 or n=2k+1n=2k+1, where kk is an integer.
Observe that if you’re given an even integer, that even integer is always in between two odd integers. For instance, the even integer 44 is between 33 and 55.
You can take it apart. There are a top and bottom (both the same) right triangle. So you can find the area of that by multiplying 8*6 and divide by two. Then multiply by two because there are 2 triangles.
You are left with three rectangular sides: One 10x10, one 10x6, and one 10x8.
So your whole equation looks like this: A = 2[(8*6)/2]+(10*10)+(10*6)+(10*8)
Answer:
Step-by-step explanation:
1). Step 4:
![x=\sqrt[3]{5^4}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E4%7D)
[Since, ![a^{\frac{1}{3}}=\sqrt[3]{a}](https://tex.z-dn.net/?f=a%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Ba%7D)
]
![x=\sqrt[3]{5\times 5\times 5\times 5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5Ctimes%205%5Ctimes%205%5Ctimes%205%7D)
Step 5:
![x=\sqrt[3]{(5)^3\times 5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%285%29%5E3%5Ctimes%205%7D)
![x=\sqrt[3]{5^3}\times \sqrt[3]{5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E3%7D%5Ctimes%20%5Csqrt%5B3%5D%7B5%7D)
2). He simplified the expression by removing exponents from the given expression.
3). Let the radical equation is,
Step 1:
Step 2:
Step 3:
Step 4:
4). By substituting in the original equation.
There is no extraneous solution.
