3 years ago

Help needed please!!!!!

Mathematics

1 answer:

larisa [96]3 years ago

3 0

Answer:

I cheated and used a software lol.

Step-by-step explanation:

Hope this helps.

You might be interested in

Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

6 0

3 years ago

Help with this please!!!!!!!!!!!!!!

sveta [45]

Answer:

A is the correct answer :)

8 0

3 years ago

Read 2 more answers

Please help me i don't know what else to put here.........................

Pepsi [2]

ANSWERRR
1. 15%
2.20%
3.45%

3 0

2 years ago

HELP ASAPPPPPPPPPPPPPPPPP

eimsori [14]

Answer:

b. Kilograms

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Donna rolled 2 number cubes each labeled 1 to 6. After each roll, she found the product of the two numbers rolled. If Donna does

a_sh-v [17]

C. 125

Step by step exploitation:

5 0

3 years ago