What is the area of the shaded sector of Point B below?

Answer: If I did it correctly it should be 150.8

Step-by-step explanation:

The formula for the volume of a cylinder is height x π x (diameter / 2)2, where (diameter / 2) is the radius of the base (d = 2 x r), so another way to write it is height x π x radius2.

Hope this helps!

Answer:

  32 pounds

Step-by-step explanation:

Pounds of potatoes can be found by multiplying pounds of onions by 5 1/3:

  (5 1/3)(6 pounds) = 32 pounds

The restaurant bought 32 pounds of potatoes.

Answer:

It's a fraction

Step-by-step explanation:

You have to figure out the meaning of the letters and find what goes on top

[tex]y=-x+5~~~~~.....(i)\\\\x-y=1~~~~~~.....(ii)\\\\\\(i)-(ii):\\\\~~~~~y-x+y=-x+5-1\\\\\implies 2y=4\\\\\implies y= \dfrac 42\\\\\implies y=2\\\\\text{Substitute y=2 in eq (ii):}\\ \\~~~~ x-2 = 1\\\\\implies x= 1+2\\\\\implies x = 3\\\\\\\text{Hence,}~~ (x,y) = (3,2)[/tex]

Answer:

[tex]x = 52[/tex]

Step-by-step explanation:

Step 1:  Create an expression

[tex]2x + 76 = 180[/tex]

Step 2:  Subtract 76 from both sides

[tex]2x + 76 - 76 = 180 - 76[/tex]

[tex]2x = 104[/tex]

Step 3:  Divide both sides by 2

[tex]\frac{2x}{2} = \frac{104}{2}[/tex]

[tex]x = 52[/tex]

Answer: [tex]x = 52[/tex]

Answer:

x = 52

Step-by-step explanation:

the 2 inside angles add up to 180 degress since they lie on the same side on the transversal line. so we set the eqaution up:

2x+76=180

2x=104

x=54

Step-by-step explanation:

the length and the width of a rectangle create a right-angled triangle, with the diagonal being the Hypotenuse of that triangle (the baseline, opposite of the 90° angle).

so, we can use Pythagoras

c² = a² + b²

with c being the Hypotenuse.

so, we have here

diagonal² = 40² + 9² = 1600 + 81 = 1681

diagonal = sqrt(1681) = 41 in

Answer:

See below

Step-by-step explanation:

sf = s0 + v0 t  + 1/2 at^2

        1  + (-3)t + 1/2 (.2t) t^2           ( weird acceleration value)

s =  1 -3t + .1 t^3           You will need a 't' value to determine the position

Answer:

B. 20πb³ + 12πb²

Step-by-step explanation:

The equation for the volume is incorrect, the volume of a cylinder with radius (r), and height (h), is πr²h.

The volume of the cylinder in terms of[tex]20\pi b^3 + 12\pi b^2[/tex]. Option B is correct.

A cylinder is a three-dimensional figure with two bases that are joined with a curved surface.

The total space occupied by the three-dimensional figure is called volume.

Given that:

The Volume is V = [tex]\pi r^2 h[/tex]

Radius r =  2b

Height,h = 5b +3

Substitute the values into the formula for the volume of a cylinder

V = πr²h

V =[tex]\pi \times (2b)^2 \times (5b + 3)[/tex]

Simplify the expression:

V = [tex]4\pi b^2 \times (5b + 3)[/tex]

Use the distributive property to get the values.

V = [tex]20\pi b^3+ 12\pi b^3[/tex]

So, the volume of the cylinder in terms of[tex]20\pi b^3 + 12\pi b^2[/tex]. Option B is correct.

Learn more about cylinder here:

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Answer:

The function is C(d) = 0.69d+25.95. The equation will be 0.69(57)+2595.

for the last question 0.69d+25.95 = 57, so 0.69d = 31.05, so d is 45

The max distance you can drive with 120 dollars is 45 miles.

WHEN GRAPHING

Step 1. Convert to y = mx + b

Step 2. Graph like normal

Step 3. Shade as indicated by the inequality symbol

Step 4. If  ≤ or ≥ ONLY, then also shade the line

FOR THIS PROBLEM

1. Graph each equation

2. Shade ABOVE line for each

3. Shade line first equation as well

Hope this helps and God bless!

Answer:

trigonometric function whose value for the complement of an angle is equal to the value of a given trigonometric function of the angle itself the sine is the cofunction of the cosine.

Answer:

[tex]derivative n thof \: sin2x [/tex]

Replace [tex]x\mapsto \tan^{-1}(x)[/tex] :

[tex]\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx[/tex]

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace [tex]x\mapsto\frac1x[/tex] :

[tex]\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx[/tex]

Then the original integral is equivalent to

[tex]\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx[/tex]

Recall that for |x| < 1,

[tex]\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}[/tex]

so that we can expand the integrand, then interchange the sum and integral to get

[tex]\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx[/tex]

Integrate by parts, with

[tex]u = \ln(x) \implies du = \dfrac{dx}x[/tex]

[tex]du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}[/tex]

[tex]\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)[/tex]

Recall the Fourier series we used in an earlier question [27217075]; if [tex]f(x)=\left(x-\frac12\right)^2[/tex] where 0 ≤ x ≤ 1 is a periodic function, then

[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}[/tex]

[tex]\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)[/tex]

[tex]\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)[/tex]

Evaluate f and its Fourier expansion at x = 1/2 :

[tex]\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)[/tex]

[tex]\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)[/tex]

[tex]\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}[/tex]

So, we conclude that

[tex]\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}[/tex]

Answer:

Step-by-step explanation: yes, we belive that they start young with pyromania.  They may see as something powerful

The given equation represents a hyperbola. Its main features are:

A hyperbola can be defined by its center, vertices,  foci and asymptotes. And it is represented algebraically  by the standard equation:  [tex]\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1[/tex], where:

h= x-coordinate of center

k= y-coordinate of center

a and b= semi-axis

First, you need to rewrite the given equation 9y²-x²+2x+54y+62=0 in the standard equation hyperbola:

[tex](-x^2+2x+?)+9(y^2+6y+?)=-62\\(-x^2+2x+?)+9(y^2+6y+9)=-62\\(-x^2+2x-1)+9(y^2+6y+9)=-62\\(-x^2+2x-1)+9(y^2+6y+9)=-62-1+81\\((-x^2+2x-1)+9(y^2+6y+9)=18 ) \div 18\\ \frac{(-x^2+2x-1)}{18} +\frac{9(y^2+6y+9}{18}= \frac{18}{18} \\\frac{-(x-1)^2}{18} +\frac{(y+3^2)}{2}=1\\\frac{-(x-1)^2}{(3\sqrt{2})^2} +\frac{(y+3^2)}{(\sqrt{2})^2 }=1\\\frac{\left(y+3)^2}{\left(\sqrt{2}\right)^2}-\frac{\left(x-1\right)^2}{\left(3\sqrt{2}\right)^2}=1[/tex]

Comparing the previous equation with the standard form ([tex]\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1[/tex]), you have:

h=1,  k=-3,  a=[tex]\sqrt{2}[/tex] and b=[tex]3\sqrt{2}[/tex] . From now, it is possible to find that the question asks:

The coordinates for center is (h,k). Thus, the center is (1,-3).

The vertices (V1 and V2) of hyperbola can be found from the coordinates of center (h,k) and the semi-axis (a).

V1= (h,k+a)= (1,-3+[tex]\sqrt{2}[/tex])

V2= (h,k-a)= (1,-3-[tex]\sqrt{2}[/tex])

The foci or the focus points can be found from the coordinates of center (h,k) and the c ([tex]\sqrt{a^2+b^2}[/tex]) which represents the distance from the center to the focus.

[tex]c=\sqrt{a^2+b^2}\\ c=\sqrt{(3\sqrt{2} )^2+(\sqrt{2} )^2}\\c=\sqrt{18+2} \\c=\sqrt{20}=2\sqrt{5}[/tex]

Thus,

F1= (h,k+c)= (1 , -3+ [tex]2\sqrt{5}[/tex] )

F2= (h,k-c)= (1 , -3 - [tex]2\sqrt{5}[/tex] )

The asymptotes are the lines the hyperbola tends to at ±∞. For hyperbola, the asymptotes are defined as: [tex]y=\pm \frac{a}{b}\left(x-h\right)+k[/tex]. Then, for this question:

[tex]y=\pm \frac{\sqrt{2}}{3\sqrt{2}}\left(x-1\right)-3\\y=\pm \frac{1}{3}\left(x-1\right)-3[/tex]

Read more about hyperbola here:

https://brainly.com/question/3351710

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Answer:

yes

Step-by-step explanation:

yes

each input (x) has exactly one output (y)

in other words

it passes the vertical line test

Answer:

[tex]\large\lim_{x\rightarrow 1} f\left( x\right)=8[/tex]

Step-by-step explanation:

[tex]\lim_{x\rightarrow 1} \frac{f\left( x\right) -8}{x-1} =10[/tex]

[tex]\lim_{x\rightarrow 1} f\left( x\right) -8=10\times \lim_{x\rightarrow 1}\left( x-1\right)[/tex]

[tex]\lim_{x\rightarrow 1} f\left( x\right) -8=10\times\left( (1)-1\right)[/tex]

[tex]\lim_{x\rightarrow 1} f\left( x\right) -8=10\times\left( 0\right)=0[/tex]

[tex]\lim_{x\rightarrow 1} f\left( x\right) -8=0[/tex]

[tex]\lim_{x\rightarrow 1} f\left( x\right) =8[/tex]

Answer:

3

Step-by-step explanation:

((7x-4)^2) - (7x-4)(7x-4) + 3

(7x-4)(7x-4) is 7x-4^2

So its (7x-4)^2 - (7x-4)^2 + 3

(7x-4)^2 cancel out.

Answer is 3

Answer:

1.33 hours(If it doesn't let you type that in, just do 1 hour for rounding purposes)

Step-by-step explanation:

First, we take the 350 gal variable and divide that by 4.39, we now get 79.73

Now, we convert that into hours(60mins per hour), so 79.73 divided by 60.

We get 1.33, which is your answer!

The student T distribution would be used since we don't know the population variance (by extension the population standard deviation).

However, if n > 30, then the student T distribution is going to look very similar to the standard normal Z distribution. It won't be a perfect match, but it'll be close enough that it's more simple to use the Z distribution. With the Z distribution, you only have one set of values and you don't have to worry about the degrees of freedom.

Answer:

see explanation

Step-by-step explanation:

under a reflection in the x- axis

a point (x, y ) → (x, - y ) , then

(3, 4 ) → (3, - 4 )

(3, 1 ) → (3, - 1 )

(5, 1 ) → (5, - 1 )

Answer:

Points would be (3,-1) (5,-1) (3,-4)

Step-by-step explanation:

So its asking for a reflection on the x-axis, which means you simply take the shape and move the points to the same spot on the other side of the boldest horrizontal line, or in other words, you can just add a negative to the y variable in the points, because the rule is (x,-y).

Answer:

1992

Step-by-step explanation:

60x38=2280

32x9=288

2280-288=1992

Hope it helps!

Answer:

x=1 y=2

Step-by-step explanation:

Isolate x for 4x -y =2: x=2+y/4

Substitute x= 2+y/4

(2+13y/4) +3y=7

simplify 2+13y/4) =7

Isolate y for 2+13y/4) =7: y=2

Forx= 2+y/4

Substitute y =2

x=2+2/4

x=1

Answer:

14

Step-by-step explanation:

g(f(1)) simply means we first get f(1), and that result becomes the input value for g(x). and that is then the end result.

so,

f(1) = 4

and then

g(f(1)) = g(4) = 14

[tex]\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{"f" varies inversely with "L"}}{f=\cfrac{k}{L}}\qquad \textit{we also know that} \begin{cases} f=\stackrel{lbs}{20}\\ L=\stackrel{ft}{1.5} \end{cases} \\\\\\ 20=\cfrac{k}{1.5}\implies 30=k~\hfill \boxed{f=\cfrac{30}{L}} \\\\\\ \textit{when L = 6, what is "f"?}\qquad f=\cfrac{30}{6}\implies f=5[/tex]

Answer:

Step-by-step explanation:

Answer:

it is 57

Step-by-step explanation:

Answer:

A. √57

Step-by-step explanation:

√47 would be off the number line because it's below the number 7

√65 would be past 8 but the red dot comes before the number 8

√72 would be past 8 as well and it would be past √65 but the dot is before number 8

This leaves our answer as √57 beause the square root of 57 is 7.549 (etc.) which is between the numbers 7 and 8, while also being slightly higher than 7.5 :)

Have a good day!