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solong [7]
4 years ago
10

Find the zeros for function x^4-4x

Mathematics
1 answer:
chubhunter [2.5K]4 years ago
3 0

Factor x4−4x

x4−4x

=x(x3−4)

Answer:

x(x3−4)

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A recursion formula and the initial term of a sequence are given. Write out the first five terms of the sequence. a Subscript fo
Lelu [443]

Answer:

6, -6, 6, -6 and 6.

Step-by-step explanation:

Given the recursion formula for a sequence

a_{n+1}=-a_n\\$where a_1=6\\

The first five terms of the sequence are:

\text{First Term, }a_1=6\\$Second Term, a_2=a_{1+1}=-a_1=-6\\$Third term, a_3=a_{2+1}=-a_2=6\\$Fourth term, a_4=a_{3+1}=-a_3=-6\\$Fifth term, a_5=a_{4+1}=-a_4=6

Therefore, the first five terms of the sequence:

a_1,a_2,a_3,a_4,a_5=6, -6, 6, -6$ and 6.

4 0
3 years ago
PLEASE HELP 80 POINTS!!
motikmotik

Answer:

  Bagel

                                       plain                          

Tuna Eggs Chicken salad Cream cheese butter

                                      Poppy

Tuna Eggs Chicken salad Cream cheese butter

                                      Wheat

Tuna Eggs Chicken salad Cream cheese butter

3b

There are 15 combinations because tuna, eggs, chicken salad, cream cheese, and butter = 5

there are 3 different types of bagels plain poppy and wheat. with this information you can tell that the 5 combinations x 3 bagel types equal 15 total

                                   

7 0
4 years ago
Areal estate agent makes a 6% commission on the sale of a house.What is the commission on a house that costs $225,000?
Softa [21]

Answer:

$13,500

Step-by-step explanation:

convert percentage to decimal first by dividing it by 100. Multiply the commission percentage by the purchase price to find out your total commission.

8 0
3 years ago
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natima [27]

Answer:

Step-by-step explanation:

graph_1 and graph_2 are for this task

graph_3 is for previous task.

8 0
3 years ago
if a and b are the two odd positive integers such that a>b then one of the two numbers ( a+b) /2 and (a-b) /2is odd and even
Tamiku [17]

If a and b are odd positive integers, they are one more than a non-negative even number, i.e. there exists m,n \in \mathbb{N} such that

a = 2m+1,\quad b = 2n+1

So, the first expression become

\cfrac{a+b}{2} = \cfrac{2m+1+2n+1}{2} = \cfrac{2m+2n+2}{2} = m+n+1

Similarly, we have

\cfrac{a-b}{2} = \cfrac{2m+1-2n-1}{2} = \cfrac{2m-2n}{2} = m-n

Now, the parity of these expressions depend on those of m and n. We have four cases:

If both m and n are even:

m+n+1 is odd, since m+n is even, while m-n is even

If one of the two is odd and the other is even:

m+n+1 is even, since m+n is odd, while m-n is odd

If both are odd:

m+n+1 is odd, since m+n is even, while m-n is even

So, in all cases, one between (a+b)/2 and (a-b)/2 is odd, and the other is even.

7 0
3 years ago
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