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2 years ago

Pls help asap solve each inequality 9n - 1 < 10n 5h ≤ 12 + 4h

Mathematics

2 answers:

katrin2010 [14]2 years ago

8 0

Sorry for wasting your time the app told me to answer 1 question

Stella [2.4K]2 years ago

6 0

Here try this answer

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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

2 years ago

Give the scale factor from small to large for the similar figures

V125BC [204]

The scale factor is 5.

If you multiply each side of the small triangle by 5, you get the length of the large triangle

Hope this helps

4 0

3 years ago

Can someone help me please? It’s like my 3rd time trying to get some answers for this question. Thanks. :)

mote1985 [20]

Answer:

both rates of change equal the slope of the line (3/4)

Step-by-step explanation:

Part a)

We calculate the rate of change using the formula:

$\frac{y_2-y_1}{x_2-x_1}$

for the first interval [0,6], we calculate the y-values at x=0 and x=6;

at x=0 : $y=\frac{3}{2} (0)+2=2$

at x=6 : $y=\frac{3}{2} (6)+2=11$

therefore, the rate of change in this interval is: $\frac{y_2-y_1}{x_2-x_1}=\frac{11-2}{6-0}=\frac{9}{6} =\frac{3}{2}$

For the second interval [-4,4], we calculate the y-values at x=-4 and x=4;

at x=-4 : $y=\frac{3}{2} (-4)+2=-4$

at x=4 : $y=\frac{3}{2} (4)+2=8$

therefore, the rate of change in this interval is: $\frac{y_2-y_1}{x_2-x_1}=\frac{8-(-4)}{4-(-4)}=\frac{12}{8} =\frac{3}{2}$

Part b):

Notice that both rates of change equal the value of the slope of the linear function (3/2)

4 0

2 years ago

Is this correct? <br>Pls Help!!

lubasha [3.4K]

Answer:

if it equals 3276 then yes its correct

Step-by-step explanation:

3 0

3 years ago

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A line plot with the lengths of meteors is shown. Two more identical meteor measurements were added so that the total length of

liraira [26]

Answer: Could you give us the graph?

Step-by-step explanation:

3 0

2 years ago

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