You might want to stick to at most five questions at once, makes it easier for the rest of us. :)
17. T has a vertical line of symmetry (along the center line).
18. Z looks the same if you turn it halfway around.
19. The passes total to 150°, which is a little less than 180°, so I estimate it would be in front of Kai.
20. Left is the -x direction. Up is the +y direction. this is (x-6, y+4)
21. Every dilation has a center (where it's dilated from) and a scale factor (how much it's dilated).
22. It must be A, because it's the only one where the number of moves adds up to 16.
23. It can be determined to be B just by tracking where point C would end up through the transformation.
24. A 180° rotation flips the signs on both components to give you (-1, 6).
25. Right is the +x direction. Down is the -y direction. (x+3, y-5)
26. This is a reflection.
Need clarification on anything?
From the choices above the answer would be: D 7y^4-13x^3 inches
If the exclusion of Miss Jones and Mr Smith serving together was not present, the are 15C4 = 1365 ways of selecting the committee.
Miss Jones can serve on the committee in the following ways:
a) with 3 men
b) with 2 men and another woman
c) with 1 man and 2 other women
d) with 3 other women.
Arrangement d) obviously presents no restrictions.
Arrangement a) has 7C3 ways excluding Mr Smith, and 8C3 ways if Mr Smith was included.
Arrangement b) has 7C2 * 6 ways excluding Mr Smith, and 8C2 * 6 ways if Mr Smith was included.
Arrangement c) has 7 * 6C2 ways excluding Mr Smith, and 8 * 6C2 ways if Mr smith was included.
The reductions in ways caused by the restriction are as follows:
a) 8C3 - 7C3 = 21 ways
b) 6(8C2 - 7C2) = 42 ways
c) 6C2 = 15 ways
The total reduction in the number of ways is: 21 + 42 + 15 = 78.
Therefore the total number of ways of selecting the committee, while observing the restriction, is 1365 - 78 = 1287 ways.
Answer:
![(\sqrt[7]{3} )^{4}](https://tex.z-dn.net/?f=%28%5Csqrt%5B7%5D%7B3%7D%20%29%5E%7B4%7D)
Step-by-step explanation:
The only radical that matches the equivalent answer is the fifth one down. We can easily eliminate the radicals without exponents on the outside, since we know they won't create leftover fractions. So that leaves us with the second, fourth and fifth answers to contemplate.
Let's look at ![(\sqrt[4]{3} )^{7}](https://tex.z-dn.net/?f=%28%5Csqrt%5B4%5D%7B3%7D%20%29%5E%7B7%7D)
and ![\sqrt[4]{3^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B3%5E%7B7%7D%20%7D)
first. It's good to know that these are equivalent radicals. The numbers are the same, and they will produce the same answers.
When you do the math, the exponent rule gives us fractions of 
for exponents, and eventually, a 
for both answers. So these are eliminated.
Now, for ![(\sqrt[4]{3}^{7}](https://tex.z-dn.net/?f=%28%5Csqrt%5B4%5D%7B3%7D%5E%7B7%7D)
, we can easiy simplify by changing the 7th root to a fraction in our exponent. Use the rule: ![\sqrt[n]{x} = x^{\frac{1}{n} }](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D)
- <em>Multiply the exponents:</em>
- <em>Insert the product into the exponent: </em> <u>
</u>
And we can see the answer we're looking for! If you use this method to look at the other problems, you'll see that this is the only radical that simplifies to the required answer.
