Answer: A = 198.6cm²
Step-by-step explanation: The figure includes two semi circles on both ends. These can be solved as a whole circle with a diameter of 10cm because each semi circle is 5cm.
r = 10/2= 5cm
Area of circle:
A = πr²
A = 3.142*5²
A = 3.142*25
A = 78.55
A = 78.6cm²
The remaining section of the shape is a rectangle with a length of 12cm.
To get this, you subtract 10 from 22 because each semi circle is 5cm which adds up to 10cm.
The rectangle has a width of 10cm.
Area of rectangle:
A = lw
A = 12*10 = 120cm²
The total area of the shape:
A = 120 + 78.6
A = 198.6cm²
Answer:
A =530.66 cm^2
Step-by-step explanation:
A DVD is a circle so we can use the formula for area of a circle
A = pi r^2
We can find the radius from
r = d/2 where d is the diameter
r = 26/2
r = 13
A = pi *13^2
A = pi *169
Substituting in for pi
A = (3.14) * 169
A =530.66
Answer:
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or

Step-by-step explanation:
Given
![f(x) = \sqrt[3]x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5Dx)
Required
Write a rule for g(x)
See attachment for grid
From the attachment, we have:


We can represent g(x) as:

So, we have:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
For:

![2 = n * \sqrt[3]{-1}](https://tex.z-dn.net/?f=2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D%7B-1%7D)
This gives:

Solve for n


To confirm this value of n, we make use of:

So, we have:
![-2 = n * \sqrt[3]1](https://tex.z-dn.net/?f=-2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D1)
This gives:

Solve for n


Hence:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or:

Answer:
1
Step-by-step explanation:
Set the height of the bar to 1 because there is only 1 number between 40-49 i.e. 49
Use the arithmetic operations to get the variable x on one side of the equation and everything else on the opposite side.
If something is being is being done to a variable, we undo that operation by using the inverse of that operation.
For example, if 10 is being added to x, we use the inverse of addition or subtraction.
18 - 7x = -20.5
We variable x is being multiplied by 7 and is subtracting 18. We need to undo all those operations.
18 - 7x = -20.5
-7x = -38.5
Now the variable is only being multiplied by -7. Reverse the operation.
-7x = -38.5
x = 5.5
So, x is equal to 5.5.