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artcher [175]
4 years ago
14

Which products result in a difference of squares? Check all that apply. (5z + 3)(–5z – 3) (w – 2.5)(w + 2.5) (8g + 1)(8g + 1) (–

4v – 9)(–4v + 9) (6y + 7)(7y – 6) (p – 5)(p – 5)
Mathematics
2 answers:
Helen [10]4 years ago
4 0

Answer:

Option 2  (w-2.5)(w+2.5)=w^2-2.5^2\\ and option 4 (-4v-9)(-4v+9)=-(4v+9)(-)(4v-9)=4v^2-9^2

Explanation:

In option 1 (5z+3)(-5z-3)=-(5z+3)(5z+3)=-(5z+3)^2  which is not the difference of squares

In option 2 (w-2.5)(w+2.5)=w^2-2.5^2\\ using the formula

(a+b)(a-b)=a^2-b^2\\Here \\a=w,b=2.5

Clearly, the difference of squares

In option 3(8g+1)(8g+1)=(8g+1)^2 not the difference of squares

In option 4(-4v-9)(-4v+9)=-(4v+9)(-)(4v-9)=4v^2-9^2

Clearly the difference of squares

In option 5 (6y+7)(7y-6)=42y^2+13y-42 not the difference of squares

In option 6 (p-5)(p-5)=(p-5)^2 not the difference of squares

Sloan [31]4 years ago
3 0
<span>these are the answers because i had the same question and they told me these were the answers 
</span>(w – 2.5)(w + 2.5)<span>
</span>
<span>(–4v – 9)(–4v + 9)</span>
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Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B
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\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0

M_y=35x^4+14x^5-6y^2-12xy^2
N_x=35x^4-6y^2

\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2

This suggests an integrating factor depending on x only is possible, and given by

\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}

Distributing across the ODE, we end up with

\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0

The equation is now exact, with

{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}

Now we find the solution:

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F(x,y)=(7x^5y-2xy^3)e^{2x}=C
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