What is LE A. 3 B. 6 C. 12 D. 18

Solve x2 - 4 = 5 by graphing the related function,

Black_prince [1.1K]

It’s 3 and -3, I can explain if you want

5 0

3 years ago

Show why the limit as x approaches 0 (csc(x)-cot(x)) involves an indeterminate form, and then prove that the limit equals 0.

Nadusha1986 [10]

Answer with Step-by-step explanation:

We are given that $\lim_{x\rightarrow 0 }(csc(x)-cot(x))$

We have to prove that why the limit x approaches 0(csc(x)-cot(x)) involves an indeterminate form and prove that the limit equals to 0.

$\lim_{x\rightarrow 0 }(\frac{1}{sinx}-\frac{cosx}{sinx})$

Because $csc(x)=\frac{1}{sinx}$ and $cot(x)=\frac{cosx}{sinx}$

$\lim_{x\rightarrow 0 }(\frac{1-cosx}{sinx})$

$\frac{1-cos0}{sin0}$

We know that cos 0=1 and sin 0=0

Substitute the values then we get

$\frac{1-1}{0}=\frac{0}{0}$

We know that $\frac{0}{0}$ is indeterminate form

Hence, the limit x approaches 0(csc(x)-cot(x)) involves an indeterminate form.

L'hospital rule:Apply this rule and differentiate numerator and denominator separately when after applying $\lim_{x\rightarrow a }$ we get indeterminate form $\frac{0}{0}$

Now,using L' hospital rule

$\lim_{x\rightarrow 0 }\frac{0+sinx}{cosx}$

because $\frac{dsinx}{dx}=cosx,\frac{dcosx}{dx}=-sinx}$

Now, we get

$\lim_{x\rightarrow 0 }\frac{sinx}{cos x}$

$\frac{sin0}{cos0}$

$\frac{0}{1}=0$

Hence, $\lim_{x\rightarrow 0 }(csc(x)-cot(x))=0$

5 0

3 years ago

if a point is chosen inside the large circle what is the probability that it will also be inside the small circle?

Ira Lisetskai [31]

Answer:

0.25

Step-by-step explanation:

hope that helps! ^w^

8 0

2 years ago

Read 2 more answers

HELP PLEASE............... ............... ............... ...............

Alexus [3.1K]

Whats the question, youre just showing the answer psycho

5 0

3 years ago

Im a dude and the question is what is the length of the line segment between A(-5,8) and B(7,8)

pishuonlain [190]

$\text {Length = } \sqrt{(8-8)^2 +(7+5)^2 } = \sqrt{12^2} = 12$

Answer: 12 units

7 0

3 years ago

Read 2 more answers