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True [87]
4 years ago
5

Whats the answer to 26.02 x 2.006

Mathematics
1 answer:
finlep [7]4 years ago
8 0

Answer:

52.19612

Step-by-step explanation:

hope this helps

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8. Graph: y = x2 + 5x - 1​
vredina [299]

Answer:

boyyyy

Step-by-step explain:

yayyy

7 0
3 years ago
Read 2 more answers
a car dealer paid a certain and marked it up by 7\5 of the price he paid later he sold it for $24000 what is the original price
DerKrebs [107]
X = original price

(7/5)x = 24000

x = (5/7)24000

x = 17143

the original price was $17,143
4 0
4 years ago
What is the nth term rule of the quadratic sequence below?
Ahat [919]

Answer:

The nth term = n^2 - 2n - 3.

Step-by-step explanation:

Here, we will be finding the nth term of a quadratic number sequence. A quadratic number sequence has nth term = an² + bn + c

Example 1

Write down the nth term of this quadratic number sequence.

-3, 8, 23, 42, 65...

Step 1: Confirm the sequence is quadratic. This is done by finding the second difference.

Sequence = -3, 8, 23, 42, 65

1st difference = 11,15,19,23

2nd difference = 4,4,4,4

Step 2: If you divide the second difference by 2, you will get the value of a.

4 ÷ 2 = 2

So the first term of the nth term is 2n²

Step 3: Next, substitute the number 1 to 5 into 2n².

n = 1,2,3,4,5

2n² = 2,8,18,32,50

Step 4: Now, take these values (2n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.

n = 1,2,3,4,5

2n² = 2,8,18,32,50

Differences = -5,0,5,10,15

Now the nth term of these differences (-5,0,5,10,15) is 5n -10.

So b = 5 and c = -10.

Step 5: Write down your final answer in the form an² + bn + c.

2n² + 5n -10

Example 2

Write down the nth term of this quadratic number sequence.

9, 28, 57, 96, 145...

Step 1: Confirm if the sequence is quadratic. This is done by finding the second difference.

Sequence = 9, 28, 57, 96, 145...

1st differences = 19,29,39,49

2nd differences = 10,10,10

Step 2: If you divide the second difference by 2, you will get the value of a.

10 ÷ 2 = 5

So the first term of the nth term is 5n²

Step 3: Next, substitute the number 1 to 5 into 5n².

n = 1,2,3,4,5

5n² = 5,20,45,80,125

Step 4: Now, take these values (5n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.

n = 1,2,3,4,5

5n² = 5,20,45,80,125

Differences = 4,8,12,16,20

Now the nth term of these differences (4,8,12,16,20) is 4n. So b = 4 and c = 0.

Step 5: Write down your final answer in the form an² + bn + c.

5n² + 4n

5 0
3 years ago
Find all zeros of f(x)=x^4-3x^3+6x^2+2x-60 given that 1+3i is a zero of f(x) ​
Triss [41]

Answer:

The roots of f(x) are: -2, 3, (1+3i) and (1-3i)

Step-by-step explanation:

We are given an expression:

f(x)=x^4-3x^3+6x^2+2x-60

(1+3i) is a root of f(x)

We have to find the remaining roots of f(x).

Since, (1+3i) is a root of f(x),

x-(1+3i)

is a factor of given expression.

Now, we check if (1 - 3i) is a root of given function.

f(x)=x^4-3x^3+6x^2+2x-60\\f(1-3i)=(1-3i)^4-3(1-3i)^3+6(1-3i)^2+2(1-3i)-60\\= (28+96i)-3(-26+18i)+6(-8-6i)+2(1-3i)-60 = 0

Thus, (1-3i) is also a root of given function.

Since, (1-3i) is a root of f(x),

x-(1-3i)

is a factor of given expression.

Thus, we can write:

(x-(1+3i))(x-(1-3i))\text{ is a factor of f(x)}\\x^2-x(1-3i)+x(1+3i)+(1-3i)(1+3i)\text{ is a factor of f(x)}\\x^2-2x+10\text{ is a factor of f(x)}

Dividing f(x) by above expression:

\displaystyle\frac{x^4-3x^3+6x^2+2x-60}{x^2-2x+10} = x^2-x-6

To find the root, we equate it to zero:

x^2-x-6 = 0\\x^2 - 3x+2x-6=0\\x(x-3)+2(x-3)=0\\(x+2)(x-3) = 0\\x = -2, x = 3

Thus, the roots of f(x) are: -2, 3, (1+3i) and (1-3i)

4 0
4 years ago
3. True or False, the following 4 ordered pairs are solutions to the following equation:
svp [43]
1000 sjdhcbebebdbdbd
4 0
3 years ago
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