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3 years ago

Plz I need help plz

Mathematics

1 answer:

Mashutka [201]3 years ago

8 0

A, D, and E would have a negative product.

When multiplying two negatives together, you get a positive so B and C wouldn't work.

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4 years ago

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Determine whether the origin is included in the shaded region and whether the shaded region is above or below the line for the g

just olya [345]

Answer:

The correct option is

The origin is included in the shaded region and the shaded area is below the line.

Step-by-step explanation:

Y>3/2x + 2

Put x=0

Y= 2

(0,2)

Put y=0

X= -4/3

(-4/3,0)

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4 years ago

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tekilochka [14]

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3 years ago

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Show three ways to write the expression 5 times x algebraically.

Dahasolnce [82]

Step-by-step explanation:

1st way: 5x

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2 years ago

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How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t

Zarrin [17]

By first principles, the derivative is

$\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

The first term is eliminated, and the limit is

$\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h$

A power of $h$ in every term of the numerator cancels with $h$ in the denominator:

$\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)$

Finally, each term containing $h$ approaches 0 as $h\to0$ , and the derivative is

$y=x^n\implies y'=nx^{n-1}$

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4 years ago