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Mrac [35]
3 years ago
13

-9x(x-1)+(-x)(-3x+4)

Mathematics
1 answer:
Greeley [361]3 years ago
8 0
- 9 x * ( x - 1 ) + ( - x ) * ( - 3 x + 4 ) =
= - 9 x² + 9 x - x * ( - 3 x + 4 ) =
= - 9 x² + 9 x + 3 x² - 4 x =
= - 9 x² + 3 x² + 9 x - 4 x =
= - 6 x²  + 5 x  

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If f(x) = 3 x-2 what is f(5.9)
Korvikt [17]

3(5.9) = 17.7 \:  \:  \:  \: 17.7 - 2 = 15.7
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3 years ago
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Write a translation rule that maps point D(7,-3) onto point D'(2,5).
Sergio039 [100]
We are given D (7, -3) and D'(2, 5).

SAppy the transformation
D'(x,y) → D(x-5, y+8).
Then 
x=7 → x=7-5 = 2
y=-3 → y=-3+8 = 5

Answer: D (x-5, y+8) → D'
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−2x=x^2−6
Iteru [2.4K]

Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

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3 years ago
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Which choice shows the graph of a linear function?
ExtremeBDS [4]

Answer:

4th one

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because its a line (linear)

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Select the correct answer from each drop-down menu. if f(x)=x^2+2x-3 and g(x)=x^2-9, find (f/g)(4) and (f+g)(4).
Nataly_w [17]

Answer:

(f/g)(4) = 3

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Simply plug in 4 for x in both equations to find you answer!

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