Six country music bands and 3 rock bands are signed up to perform at an all-day festival. How many different orders can the bands play in if each band type must play in a row

Answer:

C

Step-by-step explanation:

In the relation stated in C the x values repeat. This shows that each x value doesn't have its own y value. By the definition of functions, this means that C cannot be a function

Answer:

The picture is the answer. Hope it's all correct. Good luck

Answer:

Im pretty sure its B

Step-by-step explanation:

The ratio to be used to find sink is A. JL : JK

The sine of an angle is the ratio of perpendicular to its hypotenuse.

Mathematically,

Sine of an angle = Perpendicluar/Hypotenuse

Now, it is given that a △JKL.

To find sine K we need to find the hypotenuse and Perpendicular of the triangle.

Since, Sine of an angle = Perpendicluar/Hypotenuse

So, sin K = Perpendicluar/Hypotenuse

Since K is the angle given so opposite of the angle K must be the perpendicular.

So, Perpendicular of the △JKL = JL

So, Hypotenuse of △JKL = JK or KL.

Since from the given option, it can be concluded that JK is the hypotenuse of △JKL.

Therefore, sin K = JL/JK

Thus, the ratio to be used to find sink is A. JL : JK

To learn more about ratio:

https://brainly.com/question/22285396

#SPJ2

Answer:

Step-by-step explanation:

John spent and gave to his two friends a total of

1.25 + 1.20 + 1.20 = $3.65

Money left

8.50 - 3.65 = $4.85

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

Thanks!

Mark me brainliest!

~[tex]FieryAnswererGT[/tex]~

Answer:

$4.85

Step-by-step explanation:

Given;

Little John had $8.50

Spent $1.25 on sweets

Gave 2 friends $1.20

Step 1; [Solve Spend on Sweets]

$8.50 - $1.25=7.25

Step 2; Find amount given to his 2 friend]

[Gave $1.20 to each of his 2 friends]

$1.20 x 2 = 2.40

7.25 - 2.40=4.85  

Hence, Little John have $4.85 is left

[RevyBreeze]

Answer:

RS = 32

Step-by-step explanation:

Set-up equation like this -- 2x+6+16=4x-4

Solve for x

Then plug x in for the length of RS. You should get 32 for RS.

Answer:

9.6%

Step-by-step explanation:

The function m(x) = ln x is a natural logarithm function

The minimum value is -2.30 and the maximum value is - 1.61

The function is given as:

m(x) = ln x

The domain is given as:

1/10 ≤ x ≤ 1/5

At x = 1/10, we have:

m(1/10) = ln (1/10)

m(1/10) = -2.30

At x = 1/5, we have:

m(1/5) = ln (1/5)

m(1/5) = -1.61

So, we have: -2.30 ≤ m(x) ≤ -1.61

Hence, the minimum value is -2.30 and the maximum value is - 1.61

Read more about functions at:

https://brainly.com/question/13473114

Answer:

The minimum value is -2.303 and the maximum value is -1.609.

Step-by-step explanation:

Please mark as brainliest

Answer:

1. HP and MN

2. HM and PN

Step-by-step explanation:

The parallel sides are the ones that do not touch.

HP and MN don't touch.

HM and PN also don't touch.

Answer:

C is nonlinear. D is linear

Step-by-step explanation:

Answer:

d

Step-by-step explanation:

Answer:

Step-by-step explanation:

As per the graph we see that:

So the rule is:

Correct option is B

Answer:

B. ry-axis(x, y) → (–x, y)

Step-by-step explanation:

Took the quiz, hope this helps! :)

Answer:

18

Step-by-step explanation:

Since it's an add 3 rule, you need to add 3 to 15

3+15=18

Science Center: 114 + (3 * x)

Dino Museum: 6 * x

114+(3*x) = 6*x

114+(3*39) = 6*39

114+117 = 234

231 = 234

There needs to be at least 39 students for the science center to cost less than the dino discovery museum. Hope this helps! Please mark brainliest:)

Answer:

424

Step-by-step explanation:

Answer:

Jaguars sleep to about 15-18 hours a day so multiply that by how many days are in a year.

Answer:

4th one

Step-by-step explanation:

They are both L's on either side.

Answer:

Ánswer 4th one ..I think

Answer:

SA = 2 (5) ( 4) + 2( 5)(8) + 2( 4)(8)

Step-by-step explanation:

SA = 2lw + 2lh + 2wh

We know the length is 5 and the height is 8 and the width is 4

SA = 2 (5) ( 4) + 2( 5)(8) + 2( 4)(8)

Answer:

x ≥ 27

Step-by-step explanation:

x/9 + 7 ≥ 10

Subtract 7 from each side

x/9 + 7-7 ≥ 10-7

x/9  ≥ 3

Multiply each side by 9

x/9 *9 ≥ 3*9

x ≥ 27

Answer:

X>27

Step-by-step explanation:

Your Welcome

Answer:

[tex]\displaystyle 3 + 21x[/tex]

Step-by-step explanation:

[tex]\displaystyle 3[7x + 1] \hookrightarrow \boxed{21x + 3}[/tex]

I am joyous to assist you at any time.

Answer:

10 - 10 rupees note

Step-by-step explanation:

have a good day .please mark as brainiest

Answer:

Diameter: 10 inch

Radius: 5 inch

Area: 78.54 inch

I hope I'm right let me know

Answer:

Step-by-step explanation:

Step-by-step explanation:

I don't see answer options, but I can very well guess what went wrong :

he used the area formula to calculate the perimeter.

the area would be length × width, in our case

6 × 5 = 30 units²

but the perimeter is 2×length + 2×width, in our case

2×6 + 2×5 = 12 + 10 = 22 units

Hello!

P= 6+5+6+5 =22u

or

2(6+5) =  2*11 => 22u

6*5 = 30 is area

Answer:

The value of the coin when Consuelo originally purchased it is $9.74 ⇒ C

Step-by-step explanation:

The form of the linear equation is;

y = m x + b, where

∵ The value of a collectible coin can be represented by the equation

   y = 2 x + 9.74

→ Compare it with the form of the linear equation above

∴ m = 2

∴ b = 9.74

∵ x represents the number of years that Consuelo has owned the coin

∵ y represents the total value, in dollars, of the coin

→ That means the value of y at x = 0 is the value of the coin when

   Consuelo originally purchased it

∵ b represents the value of y at x = 0

∵ b = 9.74

∴ The value of the coin when Consuelo originally purchased it is $9.74

Answer:

c

Step-by-step explanation:

Answer:

0.1 as a fraction is 1/10

Step-by-step explanation:

Answer:

the answer is 1/10•••••••••

Answer:

  C.

Step-by-step explanation:

Similar triangles have congruent corresponding angles, and proportional corresponding sides.

The only angle given is angle A, which will correspond to angle D if the similarity statement is true. That is, angle D must be 35°. (Eliminates choice D)

__

The ratios of the given sides are AB:AC = 9:12 = 3:4. The corresponding sides would be DE:DF. (Eliminates choice A)

The remaining choices have side ratios of ...

  B. DE : DF = 16 : 21 ≠ 3 : 4

  C. DE : DF = 12 : 16 = 3 : 4

The set of measurements that would make ΔABC ~ ΔDEF is ...

  DE = 12, DF = 16, and ∠D = 35°

Answer: It will cost Estelle roughly $8.86

Step-by-step explanation:

Answer:

48 because when dividing, the negatives cancel out making it a positive. and the parentheses don't really matter.

Answer:

Multiply 8 by 6. It's 48.

The negatives cancel out when multiplying and dividing. However, it's different when adding or subtracting.

[tex]8*6=48[/tex]

I use complex analysis to compute the integrals in question.

First, notice that the first integrand is even:

[tex]\dfrac{\cos(-x)}{(-x)^2+1}=\dfrac{\cos(x)}{x^2+1}[/tex]

[tex]\implies\displaystyle\int_0^\infty\frac{\cos(x)}{x^2+1}\,dx=\frac12\int_{-\infty}^\infty\frac{\cos(x)}{x^2+1}\,dx[/tex]

Consider a contour C that's the union of

• Γ, a semicircle of radius R in the upper half-plane, and

• the line segment connecting the points (-R, 0) and (R, 0)

On Γ, we have [tex]z=Re^{it}[/tex] with 0 ≤ t ≤ π.

Consider the complex function

[tex]f(z)=\dfrac{e^{iz}}{z^2+1}[/tex]

and notice that our original integrand is the real part of f(z). Then the integral of f(z) over C is

[tex]\displaystyle\int_Cf(z)\,dz=\lim_{R\to\infty}\left(\int_{-R}^Rf(z)\,dz+\int_\Gamma f(z)\,dz\right)[/tex]

As R → ∞, the first integral on the right is exactly twice the one we want. Estimate the second one to be bounded by

[tex]\displaystyle\left|\int_\Gamma f(z)\,dz\right|\le\pi R|f(z)|\le\frac{\pi R}{R^2-1}[/tex]

since

[tex]|z^2+1|\ge\bigg||z^2|-|-1|\bigg|=|R^2-1|[/tex]

and so the integral along Γ vanishes.

f(z) has only one pole in the interior of C at z = i. By the residue theorem,

[tex]\displaystyle\int_Cf(z)\,dz=2\pi i\,\mathrm{Res}\left(f(z),z=i\right)=2\pi i\lim_{z\to i}(z-i)f(z)=\frac\pi e[/tex]

[tex]\implies\displaystyle\int_0^\infty\frac{\cos(x)}{x^2+1}\,dx=\frac12\mathrm{Re}\left(\int_Cf(z)\,dz\right)=\frac\pi{2e}[/tex]

For the second integral, we recall that for complex z,

[tex]\sqrt z=\exp\left(\dfrac12\left(\ln|z|+i\arg(z)\right)\right)[/tex]

Consider a keyhole contour C, the union of

• [tex]\Gamma_R[/tex], the larger circle with radius R and [tex]z=Re^{it}[/tex], with 0 < t < 2π ;

• [tex]\Gamma_\varepsilon[/tex], the smaller circle with radius ε and [tex]z=\varepsilon e^{-it}[/tex], with 0 < t < 2π ;

• [tex]\ell_1[/tex], the line segment above the positive real axis joining [tex]\Gamma_\varepsilon[/tex] to [tex]\Gamma_R[/tex] ; and

• [tex]\ell_2[/tex], the other line segment below the positive real axis joining [tex]\Gamma_R[/tex] to [tex]\Gamma_\varepsilon[/tex]

Then

[tex]\displaystyle\int_Cf(z)\,dz=\int_{\Gamma_R}f(z)\,dz+\int_{\ell_1}f(z)\,dz+\int_{\Gamma_\varepsilon}f(z)\,dz+\int_{\ell_2}f(z)\,dz[/tex]

and in the limit, the integral over [tex]\ell_1[/tex] converges to the one we want.

Estimate the integrals over the circular arcs:

• [tex]\Gamma_R[/tex] :

[tex]\displaystyle\left|\int_{\Gamma_R}f(z)\,dz\right|\le2\pi R|f(Re^{it})|\le\dfrac{2\pi R^{3/2}}{|R-\sqrt5|^2}\to0[/tex]

as R → ∞.

• [tex]\Gamma_\varepsilon[/tex] :

[tex]\displaystyle\left|\int_{\Gamma_\varepsilon}f(z)\,dz\right|\le2\pi \varepsilon|f(\varepsilon e^{-it})|\le\dfrac{2\pi\varepsilon^{3/2}}{|\varepsilon-\sqrt5|^2}\to0[/tex]

as ε → 0.

Consider the integral over [tex]\ell_2[/tex] :

[tex]\displaystyle\int_{\ell_2}f(z)\,dz=\int_R^\varepsilon\frac{\sqrt z}{z^2+2z+5}\,dz\\\\=\int_R^\varepsilon\frac{\exp\left(\dfrac12\left(\ln|z|+2\pi i\right)\right)}{z^2+2z+5}\,dz\\\\=-\int_R^\varepsilon\frac{\exp\left(\dfrac12\ln|z|\right)}{z^2+2z+5}\,dz\\\\=\int_\varepsilon^R\frac{\sqrt z}{z^2+2z+5}\,dz\\\\=\int_{\ell_1}f(z)\,dz[/tex]

so in fact,

[tex]\displaystyle\int_Cf(z)\,dz=2\int_0^\infty\frac{\sqrt x}{x^2+2x+5}\,dx[/tex]

By the residue theorem,

[tex]\displaystyle\int_Cf(z)\,dz=2\pi i\sum_{\rm poles}\mathrm{Res}\,f(z)[/tex]

We have poles at z = -1 + 2i and z = -1 - 2i. On our chosen branch,

[tex]\sqrt{-1+2i}=i\sqrt[4]{5}\exp\left(-\dfrac i2\tan^{-1}(2)\right)[/tex]

[tex]\sqrt{-1-2i}=i\sqrt[4]{5}\exp\left(\dfrac i2\tan^{-1}(2)\right)[/tex]

The residues are

[tex]\mathrm{Res}(f(z),z=-1-2i)=\dfrac{i\sqrt[4]{5}\exp\left(\frac i2\tan^{-1}(2)\right)}{-4i}[/tex]

[tex]\mathrm{Res}(f(z),z=-1+2i)=\dfrac{i\sqrt[4]{5}\exp\left(-\frac i2\tan^{-1}(2)\right)}{4i}[/tex]

Their sum is

[tex]\displaystyle\sum_{\rm poles}\mathrm{Res}\,f(z)=-\frac{\sqrt[4]{5}}2\sin\left(\dfrac12\tan^{-1}(2)\right)=-\frac{\sqrt[4]{5}}2\sqrt{\frac{5-\sqrt5}{10}}=-\frac i2\sqrt{\frac1\phi}[/tex]

where ɸ = (√5 + 1)/2 is the golden ratio, and so the overall integral is

[tex]\displaystyle\int_0^\infty\frac{\sqrt x}{x^2+2x+5}\,dx=\frac\pi2\sqrt{\frac1\phi}[/tex]

Lastly, recall

[tex]\displaystyle\sum_{k=0}^\infty\frac1{k!}=e[/tex]

Then our expression reduces to

[tex]\left(\dfrac{e\times\frac\pi{2e}}{\frac\pi2\sqrt{\frac1\phi}}\right)^2=\boxed{\phi}[/tex]