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frozen [14]
3 years ago
8

PLEASE HELP ASAP will mark brainliest

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
4 0
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By definition, the determinant equals ad - bc. What is the value of when x = -2 and y = 3? I NEED HELP PLEASE
sergiy2304 [10]

Answer:

<h3>B. -84</h3>

Step-by-step explanation:

Taking the determinant of the matrices we will have;

= 4x(5y) - 2x(3y)

= 20xy - 6xy

= 14xy

Given x = -2 and y = 3

Determinant = 14(-2)(3)

Determinant = -28*3

Determinant = -84

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Add a point at (0.-2) then go down 3 and put another point connect the dots thats it

Step-by-step explanation:

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nikklg [1K]

Answer:

226.2

Im pretty sure.

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3 years ago
Find a power series representation for the function. (Assume a&gt;0. Give your power series representation centered at x=0 .)
melamori03 [73]

Answer:

Step-by-step explanation:

Given that:

f_x = \dfrac{x^2}{a^7-x^7}

= \dfrac{x^2}{a^7(1-\dfrac{x^7}{a^7})}

= \dfrac{x^2}{a^7}\Big(1-\dfrac{x^7}{a^7} \Big)^{-1}

since  \Big((1-x)^{-1}= 1+x+x^2+x^3+...=\sum \limits ^{\infty}_{n=0}x^n\Big)

Then, it implies that:

\implies  \dfrac{x^2}{a^7} \sum \limits ^{\infty}_{n=0} \Big(\Big(\dfrac{x}{a} \Big)^{^7} \Big)^n

= \dfrac{x^2}{a^7} \sum \limits ^{\infty}_{n=0} \Big(\dfrac{x}{a} \Big)^{^{7n}}

= \dfrac{x^2}{a^7} \sum \limits ^{\infty}_{n=0} \Big(\dfrac{x^{7n}}{a^{7n}} \Big)}

\mathbf{=  \sum \limits ^{\infty}_{n=0} \dfrac{x^{7n+2}}{a^{7n+7}} }}

7 0
3 years ago
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