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Illusion [34]
3 years ago
13

F(x) = (x-3) (x+8) (x-11) What are the zeros of the polynomial?

Mathematics
1 answer:
aalyn [17]3 years ago
6 0
<span>hello :
</span>F(x) = (x-3) (x+8) (x-11)
F(x) =0
<span>x-3= 0 or x+8=0 or x-11 = 0
x=3 or x=-8 or x=11
</span><span>the zeros of the polynomialare : 3   , -8   , 11 </span><span>
</span>
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At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the firs
rewona [7]

Answer:

4032 different tickets are possible.

Step-by-step explanation:

Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.

To find : How many different tickets are possible ?

Solution :

In the first race there are 9 ways to pick the winner for first and second place.

Number of ways for first place - ^9C_1=9

Number of ways for second place - ^8C_1=8

In the second race there are 8 ways to pick the winner for first and second place.

Number of ways for first place - ^8C_1=8

Number of ways for second place - ^7C_1=7

Total number of different tickets are possible is

n=9\times 8\times 8\times 7

n=4032

Therefore, 4032 different tickets are possible.

8 0
3 years ago
How do i find the answer to my question…
Aleksandr-060686 [28]

Answer:

1. 15

2. 27

3. 4

4. 72

5. 2

6. 9

7. 1

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
dayo read 4/5th of a story book which has 500 pages. how many pages of the book is not yet read by dayo?
Nutka1998 [239]
1- 4/5=1/5; 1/5 •500=500/5=100
5 0
3 years ago
The sum of two numbers is 87. if three times the smaller number is subtracted from the larger number the result is seven. find t
Stolb23 [73]
\left \{ {{a+b=87} \atop {a-3b=7}} \right.&#10;&#10;

a-3b=7
a=7+3b

a+b=87
(7+3b)+b=87
7+4b=87
4b=87-7
4b=80
b=20

a=7+3b
a=7+3*20
a=7+60
a=67
5 0
3 years ago
You have $36,948.61 in a brokerage account, and you plan to deposit an additional $3,000 at the end of every future year until y
Luba_88 [7]
Current amount in account
P=36948.61

Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n

Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11

We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000

We can solve this by trial and error.


The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000.  It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.

So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.

So it takes 15.00 years to reach the goal of 280000 years.
8 0
3 years ago
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