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3 years ago

WILL GIVE BRAINLIEST PLS ANSWER

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

7 0

Answer:

Step-by-step explanation:

Check the attachment the solution of the work is given there

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Show work. This is a BC calculus problem.

stepan [7]

Answer:

Step-by-step explanation:

Remember that the limit definition of a derivative at a point is:

$\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}$

Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:

$\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}$

Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).

The answer will thus be D.

5 0

3 years ago

Who invented mathematics and what year was it invented?

ValentinkaMS [17]

it was invented in 500b.c. when there were dinosaurs <span />

8 0

3 years ago

Read 2 more answers

3) Jackson ate a chocolate bar in 2 1/2 minutes. How much of the bar did he eat

nignag [31]

Answer:

0.4 of the chocolate bar.

Step-by-step explanation:

3 0

3 years ago

Help i’m confused i don’t know what to do

VMariaS [17]

Answer:

-84i - 12i

Step-by-step explanation:

The distributive property is: a(b+c) = ab + ac

In this case, we have -6i(-14i+2)

-6i = a

-14i = b

2 = c

-6i(-14i+2) = -6i(-14i) + -6i(2)

= 84i^2 - 12i

= -84 - 12i

7 0

3 years ago

Read 2 more answers

Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.

mart [117]

Answer:

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

Step-by-step explanation:

The given expression is

$\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}$

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

$\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}$ $[\because a^ma^n=a^{m+n}]$

$\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}$

$\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}$ $[\because a^{-n}=\dfrac{1}{a^n}]$

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is $\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

5 0

4 years ago