Answer:
Step-by-step explanation:
![a^{m}*a^{n}=a^{m+n}\\\\](https://tex.z-dn.net/?f=a%5E%7Bm%7D%2Aa%5E%7Bn%7D%3Da%5E%7Bm%2Bn%7D%5C%5C%5C%5C)
![a) 3x^{2}y^{4}*2x^{6}y=3*2*x^{2+6}*y^{4+1}\\\\=6x^{8}y^{5}](https://tex.z-dn.net/?f=a%29%203x%5E%7B2%7Dy%5E%7B4%7D%2A2x%5E%7B6%7Dy%3D3%2A2%2Ax%5E%7B2%2B6%7D%2Ay%5E%7B4%2B1%7D%5C%5C%5C%5C%3D6x%5E%7B8%7Dy%5E%7B5%7D)
![b) xz^{3} * 4x^{4}z^{5}=4*x^{1+4}*z^{3+5}\\\\= 4x^{5}z^{8}](https://tex.z-dn.net/?f=b%29%20xz%5E%7B3%7D%20%2A%204x%5E%7B4%7Dz%5E%7B5%7D%3D4%2Ax%5E%7B1%2B4%7D%2Az%5E%7B3%2B5%7D%5C%5C%5C%5C%3D%204x%5E%7B5%7Dz%5E%7B8%7D)
c)
![4a^{3}b^{2}* 3a^{6}b^{5}=4*3*a^{3+6}*b^{2+5}\\\\=12a^{9}b^{7}](https://tex.z-dn.net/?f=4a%5E%7B3%7Db%5E%7B2%7D%2A%203a%5E%7B6%7Db%5E%7B5%7D%3D4%2A3%2Aa%5E%7B3%2B6%7D%2Ab%5E%7B2%2B5%7D%5C%5C%5C%5C%3D12a%5E%7B9%7Db%5E%7B7%7D)
![d) 6s^{5}t*s^{4}t^{2}=6*s^{5+4}*t^{1+2}\\\\=6s^{9}t^{3}\\](https://tex.z-dn.net/?f=d%29%206s%5E%7B5%7Dt%2As%5E%7B4%7Dt%5E%7B2%7D%3D6%2As%5E%7B5%2B4%7D%2At%5E%7B1%2B2%7D%5C%5C%5C%5C%3D6s%5E%7B9%7Dt%5E%7B3%7D%5C%5C)
The sample space of an event is the list of possible elements of the event.
The set elements are:
- Ac = {x : 0, 5 ≤ x ≤ 10}
- A n B = {x : 3 ≤ x ≤ 4}
- A ∪ B = {x : 0 < x ≤ 7}
- A∩Bc = {x : 1 ≤ x ≤ 2}
- A^c ∪ B = {x : 0, 3 ≤ x ≤ 10}
<h3>How to determine the intervals of the subsets</h3>
The given parameters are:
S = {x : 0 ≤ x ≤ 10}
A = {x : 0 < x < 5}
B = {x : 3 ≤ x ≤ 7}
<u>(a) Ac </u>
This represents the list of elements in the universal set not in set A.
So, we have:
Ac = {x : 0, 5 ≤ x ≤ 10}
<u>(b) A ∩ B </u>
This represents the list of common elements in sets A and set B.
So, we have:
A n B = {x : 3 ≤ x ≤ 4}
<u>(c) A ∪ B </u>
This represents the list of all elements in sets A and set B, without repetition.
So, we have:
A ∪ B = {x : 0 < x ≤ 7}
<u>d) A∩Bc </u>
Given that:
B = {x : 3 ≤ x ≤ 7}
So, we start by calculating B^c i.e. the list of elements in the universal set not in set B.
So, we have:
Bc = {x : 1, 2, 8 ≤ x ≤ 10}
A∩Bc would then represent the list of common elements in sets A and set Bc
So, we have:
A∩Bc = {x : 1 ≤ x ≤ 2}
<u>(e) A^c ∪ B</u>
In (a), we have:
Ac = {x : 0, 5 ≤ x ≤ 10}
Given that:
B = {x : 3 ≤ x ≤ 7}
A^c ∪ B would then represent the list of all elements in sets Ac and set B
So, we have:
A^c ∪ B = {x : 0, 3 ≤ x ≤ 10}
Read more about sets are:
brainly.com/question/2193811
Answer:
65
Step-by-step explanation:
let the son's age be x then the man's age is 5x
In 13 years
son's age = x + 13 and man's age = 5x + 13 and he is 3 times as old as son
5x + 13 = 3(x + 13)
5x + 13 = 3x + 39 ( subtract 3x from both sides )
2x + 13 = 39 ( subtract 13 from both sides )
2x = 26 ( divide both sides by 2 )
x = 13 ← son's present age
5x = 5 × 13 = 65 ← man's present age
4 1/4 - 3 3/4 = 1/2
Sarah picked 1/2 pound more than Nathan
Answer:
![y = - \frac{1}{12} {x}^{2}](https://tex.z-dn.net/?f=y%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B12%7D%20%7Bx%7D%5E%7B2%7D%20)
Step-by-step explanation:
We want to find the equation of a parabola withfocus at (-3,0) and directrix y=3.
This para has its vertex at the origin and opens downward.
The equation is of the form
![{x}^{2} = - 4py](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%3D%20%20-%204py)
Where p is the distance from the vertex to the focus.
This distance is p=|0-3|=3
We substitute p=3 to obtain:
![{x}^{2} = - 4 \times 3y](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%3D%20%20-%204%20%5Ctimes%203y)
![{x}^{2} = -12y](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%3D%20-12y)
Hence the equation is
![y = - \frac{1}{12} {x}^{2}](https://tex.z-dn.net/?f=y%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B12%7D%20%20%7Bx%7D%5E%7B2%7D%20)