I need the below to answer
Answer:
K'O' is parallel to KO and its length = 5 * length of KO.
Step-by-step explanation:
The image is missing but I found a very similar problem on Brainly: brainly.com/question/7154536 I think the image is comparable even though it said KL instead of KO.
Assuming that is the case, because the larger figure was dilated using a scale factor of 5, K'O' is parallel to KO and its length = 5 * length of KO.
216 raised to the power 1/3 can also be written as cube root of 216.
Changing the given expression to radical form, we can write:
![(216)^{ \frac{1}{3} } \\ \\ = \sqrt[3]{216}](https://tex.z-dn.net/?f=%28216%29%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%0A%3D%20%5Csqrt%5B3%5D%7B216%7D%20)
216 is the cube of 6 i.e. multiply 6 three times (6 x 6 x 6), you will get 216.
So we can write the above expression as:
![\sqrt[3]{(6)^{3} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%286%29%5E%7B3%7D%20%7D%20)
The cube root and cube cancel out each other, leaving the answer equal to 6.
So, the correct answer is option B
Answer:
1. 3.14
Step-by-step explanation:
If we have

That just equals

For example, consider function

The inverse of that function is




So

If we compose the function, f(x) into f^-1(x).
We get

That proof of that.
1. Here x=f^-1(3.14)and f(3.14) cancel out so the answer is 3.14
2. We first find f(-7) which is -12,
We then find f(-12)= 5 so the answer here is 5.

<h2 />

<h3>a= (–2, 0) ; Center =(0,0)</h3>
<h3>

</h3><h3>C = (–5,0) ; Center =(0,0)</h3>

<h3>C²= a²+ b²</h3><h3>(5)²= (2)² + b²</h3><h3>b²= 25–4 —> b² = 21 </h3>





I hope I helped you^_^