Evaluating expressions k3+j

Mathematics

1 answer:

mrs_skeptik [129]2 years ago

3 0

The answer is 3k+j since 3 cannot be multiplied after k.

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What are the x- and y-intercepts of −6x+4y=96 ?

ankoles [38]

* intercepts "x" --> y=0

-6x + 4(0) = 96
-6x = 96
x = $- \frac{96}{6}$

x = 16

(16,0)

*intercepts "y" --> x=0

-6(0) + 4y = 96
4y = 96
y = $\frac{96}{4}$

y = 24

(0,24)

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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 $\frac{0}{0}$