Answer:
a) see below
b) 40x20 meters
Step-by-step explanation:
Write down what you know:
- The area of the enclosure is length*width, so
- The length of the fencing is 80 meters, so
Now we have to combine these two equations above, and get rid of y in the process.
First rewrite the second as:
Then substitute for y in the first:
b) To maximize A, find the zero of the first derivative:
So y = (80-40)/2 = 20 meters.
Answer:
32
Step-by-step explanation:
x+2x+75=180
3x=180-75
x=105÷3
x=32
angle T=x=32
Answer:
-10
Step-by-step explanation:
Substitute 5 into the equation
(-5)2
Simplify and get -10
Answer is y = x + 52
you can substitute x with x and substitute y with y
for example:
x = 2 y = 54
54 = 2 + 52
54 = 54
it’s true
Answer:
D. The work shown above is correct and ![x\times \sqrt[6]{x^{7}}](https://tex.z-dn.net/?f=x%5Ctimes%20%5Csqrt%5B6%5D%7Bx%5E%7B7%7D%7D)
may not be simplified further.
Step-by-step explanation:
The expression is given as,
![\sqrt[6]{x^{13}}=\sqrt[6]{x^{6}\times x^{7}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E%7B13%7D%7D%3D%5Csqrt%5B6%5D%7Bx%5E%7B6%7D%5Ctimes%20x%5E%7B7%7D%7D)
i.e. ![\sqrt[6]{x^{13}}=x\times \sqrt[6]{x^{7}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E%7B13%7D%7D%3Dx%5Ctimes%20%5Csqrt%5B6%5D%7Bx%5E%7B7%7D%7D)
Then,
Thus, we can see that above calculation is correct.
The right side of step 2 gives, ![\sqrt[6]{x^{6}\times x^{7}}=(x^{6})^\frac{1}{6}\times (x^{7})^\frac{1}{6}=x\times \sqrt[6]{x^{7}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E%7B6%7D%5Ctimes%20x%5E%7B7%7D%7D%3D%28x%5E%7B6%7D%29%5E%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%28x%5E%7B7%7D%29%5E%5Cfrac%7B1%7D%7B6%7D%3Dx%5Ctimes%20%5Csqrt%5B6%5D%7Bx%5E%7B7%7D%7D)
So, the work shown is correct and may not be further simplified.
Hence, option D is correct.
