If p-1/p=4,find the value of P2+1/p2​

Vertical asymptote:

A vertical asymptote is a value of x for which the function is not defined, that is, it is a point which is outside the domain of a function;

In a graphic, these vertical asymptotes are given by dashed vertical lines.

An example is a value of x for which the denominator of the function is 0.

In this graphic:

Dashed vertical lines at: [tex]x = -3, x = 7[/tex], thus, for [tex]x - (-3) = x+3[/tex] and [tex]x - 7[/tex] the denominator is zero.

Thus, the function graphed is:

[tex]F(x) = \frac{1}{(x+3)(x-7)}[/tex]

And the correct answer is given by option C.

To take a look at a problem with asymptote, you can check this item https://brainly.com/question/4084552.

The graph is for the rational function f(x) = 1/(x + 3)(x - 7).

Option C is the correct answer.

We have,

To understand the graph of the function f(x) = 1/((x + 3)(x - 7)).

Vertical Asymptotes:

The function has vertical asymptotes at the values of x for which the denominator becomes zero.

The denominator is (x + 3)(x - 7), so the vertical asymptotes occur at

x = -3 and x = 7.

Horizontal Asymptote:

The highest power of x in the denominator is x², and there is no x² term in the numerator, the function approaches 0 as x goes to positive or negative infinity.

The horizontal asymptote is y = 0.

x-Intercept:

To find the x-intercept, we set y = 0 and solve for x:

0 = 1/((x + 3)(x - 7))

Since the numerator can never be zero, the only way the fraction can be zero is if the denominator is zero:

(x + 3)(x - 7) = 0

Solving for x:

x + 3 = 0

x = -3

x - 7 = 0

x = 7

So, the x-intercepts are (-3, 0) and (7, 0).

y-Intercept:

To find the y-intercept, we set x = 0:

f(0) = 1/((0 + 3)(0 - 7)) = 1/(-3 * -7) = 1/21

The y-intercept is (0, 1/21).

Thus,

The graph is for the rational function f(x) = 1/(x + 3)(x - 7).

Learn more about functions here:

https://brainly.com/question/28533782

#SPJ6

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

[tex] \displaystyle \large{ cos \theta = - \frac{1}{2} }[/tex]

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

[tex] \displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }[/tex]

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }[/tex]

Both values are apart of the interval. Hence,

[tex] \displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }[/tex]

Second Question

Isolate sin(4 theta).

[tex] \displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }[/tex]

Rationalize the denominator.

[tex] \displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }[/tex]

The problem here is 4 beside theta. What we are going to do is to expand the interval.

[tex] \displaystyle \large{0 \leqslant \theta < 2 \pi}[/tex]

Multiply whole by 4.

[tex] \displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}[/tex]

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }[/tex]

Find Q4

[tex] \displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }[/tex]

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

[tex] \displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}[/tex]

Hence:-

For Q3

[tex] \displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }[/tex]

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

[tex] \displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }[/tex]

Therefore:-

[tex] \displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }[/tex]

Then we divide all these values by 4.

[tex] \displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }[/tex]

Let me know if you have any questions!

Answer:

the 4th option

Step-by-step explanation:

8 = 2³

so, when we convert

[tex] {8}^{x - 3} [/tex]

into a "2 to the power of" expression, we get therefore

[tex] ({2}^{3}) ^{x - 3} [/tex]

and when we have an exponent of an exponent, we can simplify by multiplying these two exponents.

3×(x-3) = 3x - 9

and therefore we get

[tex] {2}^{3x - 9} [/tex]

and by the way, we can now even easily solve this for x, as we know

[tex] {2}^{4x} = {2}^{3x + x} [/tex]

after all, because 4x = x + x + x + x = 3x + x = ...

and because we got also

[tex] {2}^{4x} = {2}^{3x - 9} [/tex]

we know that 3x + x = 3x - 9

and that gives us x = -9

Answer:

w

Step-by-step explanation:

w

Answer:

6000 hope this helps

if the question is 5,422 then the round figure is 5000

but the question is 5,821 its above 5500 will be 6000

Answer:

\I got not answer cause im da BUDDHA

But gimme brainliest squekky plssss

Answer:

[tex]c - 2 = 25 + 10 \sqrt c + c[/tex]

Step-by-step explanation:

Given

[tex]\sqrt{c-2} - \sqrt c = 5[/tex]

Required

Isolate radical, then square both sides

We have:

[tex]\sqrt{c-2} - \sqrt c = 5[/tex]

Isolate radical

[tex]\sqrt{c - 2} = 5 + \sqrt c[/tex]

Square both sides

[tex](\sqrt{c - 2})^2 = (5 + \sqrt c)^2[/tex]

[tex]c - 2 = 25 + 2 * 5 * \sqrt c + c[/tex]

[tex]c - 2 = 25 + 10 \sqrt c + c[/tex]

Answer:

B.

[tex]{ \tt{f(x) = \sqrt[3]{x + 11} }}[/tex]

let the inverse of f(x) be m:

[tex]{ \tt{m = \sqrt[3]{x + 11} }} \\ { \tt{ {m}^{3} = x + 11}} \\ { \tt{ {m}^{3} - 11 = x}} \\ { \tt{ {f}^{ - 1}(x) = {m}^{3} - 11 }}[/tex]

substitute for x in place of m:

[tex]{ \bf{f {}^{ - 1}(x) = {x}^{3} - 11 }}[/tex]

Elimination method

3x +4y ≥ 0

2x +3y ≥ 1

Multiply by 2, -3

6x +8y ≥ 0

-6x +-9y ≥ -3

Add

-1y ≥ -3

y = 3

3x + 12≥ 0

3x + ≥ -12

x = -4

answer: y = 3 x = -4

Answer:

The x-coordinate of this point would be 6.

Step-by-step explanation:

Answer:

-8, 5 , 0 , -7 , 16

Step-by-step explanation:

Given

[tex]t_n = (-1)^{n+1}(n^2 - 9)[/tex]

Required

The first five terms

When [tex]n = 1[/tex]

[tex]t_1 = (-1)^{1+1}(1^2 - 9)[/tex]

[tex]t_1 = (-1)^{2}(1 - 9)[/tex]

[tex]t_1 = -8[/tex]

When [tex]n =2[/tex]

[tex]t_2 = (-1)^{2+1}(2^2 - 9)[/tex]

[tex]t_2 = (-1)^3 * (4 - 9)[/tex]

[tex]t_2 = 5[/tex]

[tex]t_3 = (-1)^{3+1}(3^2 - 9)[/tex]

[tex]t_3 = (-1)^{4}(9 - 9)[/tex]

[tex]t_3 = 0[/tex]

[tex]t_4 = (-1)^{4+1}(4^2 - 9)[/tex]

[tex]t_4 = (-1)^5(16 - 9)[/tex]

[tex]t_4 = -7[/tex]

[tex]t_5 = (-1)^{5+1}(5^2 - 9)[/tex]

[tex]t_5 = (-1)^{6}(25 - 9)[/tex]

[tex]t_5 = 16[/tex]

So, the first five terms are: -8, 5 , 0 , -7 , 16

Answer:

supplement of 158 degree

x+158=180

x=180-158

x=22 degree.

Step-by-step explanation:

Answer:

[tex]8(2a- b)(4a^2+ 2ab+ b^2)[/tex]

Step-by-step explanation:

factor out the 8

then you have the sum/difference of cubes..

look that up SOAP: same opposite, always a plus

[tex]64a^3 -8b^3\\8(8a^3 -b^3)[/tex]

[tex]8(2a- b)(4a^2+ 2ab+ b^2)[/tex]

Answer:

x= -2

y= -1

z= -3

Step-by-step explanation:

x-y= -1 (1)

x+z= -5 (2)

y-z= 2 (3)

(1)+(3)==> x-z= 1 (4)

(4)+(2)==> 2x= -4 ==> x= -2

we replace x by its value in equation (2):

-2+z= -5 ==> z= -3

we replace z by its value in equation (3):

y-(-3)= 2 ==> y+3=2 ==> y= -1

The differential equation

[tex]y^{(4)}-n^2y'' = g(x)[/tex]

has characteristic equation

r ⁴ - n ² r ² = r ² (r ² - n ²) = r ² (r - n) (r + n) = 0

with roots r = 0 (multiplicity 2), r = -1, and r = 1, so the characteristic solution is

[tex]y_c=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}[/tex]

For the non-homogeneous equation, reduce the order by substituting u(x) = y''(x), so that u''(x) is the 4th derivative of y, and

[tex]u''-n^2u = g(x)[/tex]

Solve for u by using the method of variation of parameters. Note that the characteristic equation now only admits the two exponential solutions found earlier; I denote them by u₁ and u₂. Now we look for a particular solution of the form

[tex]u_p = u_1z_1 + u_2z_2[/tex]

where

[tex]\displaystyle z_1(x) = -\int\frac{u_2(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx[/tex]

[tex]\displaystyle z_2(x) = \int\frac{u_1(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx[/tex]

where W (u₁, u₂) is the Wronskian of u₁ and u₂. We have

[tex]W(u_1(x),u_2(x)) = \begin{vmatrix}e^{-nx}&e^{nx}\\-ne^{-nx}&ne^{nx}\end{vmatrix} = 2n[/tex]

and so

[tex]\displaystyle z_1(x) = -\frac1{2n}\int e^{nx}g(x)\,\mathrm dx[/tex]

[tex]\displaystyle z_2(x) = \frac1{2n}\int e^{-nx}g(x)\,\mathrm dx[/tex]

So we have

[tex]\displaystyle u_p = -\frac1{2n}e^{-nx}\int_0^x e^{n\xi}g(\xi)\,\mathrm d\xi + \frac1{2n}e^{nx}\int_0^xe^{-n\xi}g(\xi)\,\mathrm d\xi[/tex]

and hence

[tex]u(x)=C_1e^{-nx}+C_2e^{nx}+u_p(x)[/tex]

Finally, integrate both sides twice to solve for y :

[tex]\displaystyle y(x)=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}+\int_0^x\int_0^\omega u_p(\xi)\,\mathrm d\xi\,\mathrm d\omega[/tex]

This question is solved using relative frequency concepts, finding the following two way frequency table, with the - separating the values:

0.4290 - 0.0540 - 0.4839

0.0581 - 0.4581 - 0.5161

0.4871 - 0.5121 - 1

-------------------------------------------------------------------------------

Relative frequency:

The relative frequency of a to b is given by a divided by b.

-------------------------------------------------------------------------------

Democratic:

Total of 150 + 160 = 310 voters.

Of the 150 Democrats, 133 voted for the Democrat and 150 - 133 = 17 voted for the Republican.

The frequencies are:

[tex]\frac{133}{310} = 0.4290, \frac{17}{310} = 0.0548[/tex]

Proportion of democratic voters is:

[tex]\frac{150}{310} = 0.4839[/tex]

Thus, the first line is: 0.4290 - 0.0540 - 0.4839

-------------------------------------------------------------------------------

Republican:

Of the 160 Republicans, 142 voted for the Republican and 160 - 142 = 18 voted for the Democrat.

The frequencies are:

[tex]\frac{18}{310} = 0.0581, \frac{142}{310} = 0.4581[/tex]

The proportion of republican voters is:

[tex]\frac{160}{310} = 0.5161[/tex]

Thus, the second line is: 0.0581 - 0.4581 - 0.5161

-------------------------------------------------------------------------------

Third line:

0.4290 + 0.0581 = 0.4871

0.0540 + 0.4581 = 0.5121

0.4839 + 0.5161 = 1

Thus, the third line is: 0.4871 - 0.5121 - 1

-------------------------------------------------------------------------------

Two-way frequency table:

The two-way frequency table is:

0.4290 - 0.0540 - 0.4839

0.0581 - 0.4581 - 0.5161

0.4871 - 0.5121 - 1

A similar question is given at: https://brainly.com/question/24337228

Answer:

Democratic 133 18 151

Republican 17 142 159

Total        150 160 310

Step-by-step explanation:

Answer:

Yes the sales manager claims can be refuted based on the calculated percentages

Step-by-step explanation:

From the table attached below

Total number of customers = 200

Calculated percentages

customers that prefer whole life insurance = 90 = 90/200 = 45%

customers that prefer universal life insurance = 15 = 15/200 = 7.5%

customers that prefer Annuities = 95 = 95/200 = 47.5

Expected percentages:

whole life = 20%

Universal life = 10%

Life Annuities = 70%

Answer:

I believe the red one is the first image then the blue then the green because they show the prime sign

Step-by-step explanation:

the answer is

10² ( ten square )

step by step :

10² = 10 × 10

= 100

Answer: 100

Step-by-step explanation: 10*10=100

Answer:

Here's the answer i believe. x = y/m + 7

Answer:

i hope it will help you

Step-by-step explanation:

Working capital management is how companies are able to manage finances and continue operations.

Answer:

Yes one should consider to buy the policy as important to have insured plan that help at the time of need.

Step-by-step explanation:

Answer:

-1.87 (B)

865.93 - [968-3.34(30)] = -1.87

ED2021

Answer:

-1/3x+3 = x-1

Step-by-step explanation:

The solution is (3,-2)

Check and see if the point solves the equation

-1/3x+3 = x-1

-1/3(3) +3 = 3-1

-1+3 = 3-1

2=2 yes

Answer:

C

Step-by-step explanation:

Answer:

I believe it is (C).

This runner’s time was slower than the mean time by 0.75 standard deviations.

ED2021

Answer:

B) This runner’s time was faster than the mean time by 1.08 standard deviations.

Step-by-step explanation:

The runner is FASTER because of the negative z score

Answer:

Step-by-step explanation:

This is the Law of Cosines. Use the following formula:

[tex]5.1^2=3.3^2+3.3^2-2(3.3)(3.3)cos(x)[/tex] and simplify to

26.01 = 10.89 + 10.89 - 21.78cos(x) and a bit more to

4.23 = -21.78cos(x) and finally to

-.194214876 = cos(x) and use the 2nd button along with the cos button to find that the missing angle is 101.2 degrees

A well formatted table of the distribution is attached below :

Answer:

0.124

0.733

0.408

Step-by-step explanation:

Using the table Given :

1.) P(AP Statistics) = 90 / 725 = 0.124

2.) P(12th grade ; Precalculus or AP Statistics) = (100 + 65) / 225 = 165 /225 = 0.733

3.) P(Algebra 11 | 11th grade) = P(Algebara11 n 11th grade) / P(11th grade) = 100 / 245 = 0.408

Answer:

[tex](a+b)^{2} =a^{2}+2ab+b^{2}[/tex]

[tex](1)w^{2}+2(3)(1)w+3^{2}\\\\=(w+3)^{2}\\\\=(w+3)(w+3)[/tex]

Therefore, [tex]w^{2} +6w+9[/tex] makes a perfect squared.

Answer:

12

Step-by-step explanation:

Range is the subtraction of the largest number and the smallest number.

The largest number is: 23

The smallest number is: 11

Now subtract:

23 - 11 = 12

Hope this helped.

Answer:

the lowest is 11 and the highest is 24 then subtract it you are going to have 13