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notka56 [123]
3 years ago
13

Please help!!!

Mathematics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

c. 2a^3x^4-3a^3x^3+a^4x^2

d. 63x^5y^3-30x^2y^4-7x^3y^2

Step-by-step explanation:

For this exercise you need to remember the following:

 1. The multiplication of signs:

(+)(+)=+\\(-)(-)=+\\(-)(+)=-

2. The Product of powers property, which states that:

(a^m)(a^n)=a^{(m+n)}

3. The Distributive property:

a(b\±c)=ab\±ac

Knowing that, you can solve the multiplication of the polynomials:

c. (-2ax^2+3ax-a^2)(-a^2x^2)=\\\\=(-2ax^2)(-a^2x^2)+(3ax)(-a^2x^2)+(-a^2)(-a^2x^2)=2a^3x^4-3a^3x^3+a^4x^2

d. (6.3x^3y-3y^2-0.7x)(10x^2y^2)=\\\\=(6.3x^3y)(10x^2y^2)+(-3y^2)(10x^2y^2)+(-0.7x)(10x^2y^2)=63x^5y^3-30x^2y^4-7x^3y^2

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fomenos
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In case you are interested, if the equation said y = 12 x + 3 the slope of the line would still be 12 but the line would cross the y axis at 3. If the equation said y = 12x -4, the line would have a slope of 12 and would cross the y axis at -4.</span>
5 0
3 years ago
What two rational expressions sum to 2x+3/x^2-5x+4
Anni [7]

Answer:

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}

Step-by-step explanation:

Given the rational expression: \frac{2x + 3}{x^2 - 5x + 4}, to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

x^2 - 5x + 4

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(x^2 - 4x) - (x + 4)

x(x - 4) - 1(x - 4)

(x- 1)(x - 4)

Thus, we now have: \frac{2x + 3}{(x- 1)(x - 4)}

Step 2: Apply the concept of Partial Fraction

Let,

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}

Multiply both sides by (x - 1)(x - 4)

\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4) = (\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)

2x + 3 = A(x - 4) + B(x - 1)

Step 3:

Substituting x = 4 in 2x + 3 = A(x - 4) + B(x - 1)

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8 + 3 = A(0) + B(3)

11 = 3B

\frac{11}{3} = B

B = \frac{11}{3}

Substituting x = 1 in 2x + 3 = A(x - 4) + B(x - 1)

2(1) + 3 = A(1 - 4) + B(1 - 1)

2 + 3 = A(-3) + B(0)

5 = -3A

\frac{5}{-3} = \frac{-3A}{-3}

A = -\frac{5}{3}

Step 4: Plug in the values of A and B into the original equation in step 2

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}

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vitfil [10]
If x+2 is a factor of x^3+6x^2+kx+10, then if we replace x by -2 in the polynomial x^3-6x^2+kx+10, we must get zero:
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